CHAPTER 3 STOICHIOMETRY

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STOICHIOMETRY

3.1 One atomic mass unit is defined as a mass exactly equal to one-twelfth the mass of one carbon-12 atom. We cannot weigh a single atom, but it is possible to determine the mass of one atom relative to another

experimentally. The first step is to assign a value to the mass of one atom of a given element so that it can be used as a standard.

3.2 12.00 amu. On the periodic table, the mass is listed as 12.01 amu because this is an average mass of the naturally occurring mixture of isotopes of carbon.

3.3 The value 197.0 amu is an average value (an average atomic mass). If we could examine gold atoms individually, we would not find an atom with a mass of 197.0 amu. However, the average mass of a gold atom in a typical sample of gold is 197.0 amu.

3.4 You need the mass of each isotope of the element and each isotope’s relative abundance.

3.5 (34.968 amu)(0.7553)  (36.956 amu)(0.2447)  35.45 amu

3.6 Strategy: Each isotope contributes to the average atomic mass based on its relative abundance.

Multiplying the mass of an isotope by its fractional abundance (not percent) will give the contribution to the average atomic mass of that particular isotope.

It would seem that there are two unknowns in this problem, the fractional abundance of 6Li and the fractional abundance of 7Li. However, these two quantities are not independent of each other; they are related by the fact that they must sum to 1. Start by letting x be the fractional abundance of 6Li. Since the sum of the two abundance’s must be 1, we can write

Abundance 7Li  (1  x)

Solution:

Average atomic mass of Li  6.941 amu  x(6.0151 amu)  (1  x)(7.0160 amu)

6.941  1.0009x  7.0160

1.0009x  0.075 x  0.075

x  0.075 corresponds to a natural abundance of 6Li of 7.5 percent. The natural abundance of 7Li is (1  x)  0.925 or 92.5 percent.

3.7

6.022 1023amu The conversion factor required is

1 g

  

 

 

 

23

13.2 amu 1 g

6.022 10 amu

  

? g 2.191023g

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3.8

6.022 1023amu The unit factor required is

1 g

  

 

 

 

6.022 1023amu

8.4 g =

1 g

   24

? amu 5.110 amu

3.9 The mole is the amount of a substance that contains as many elementary entities (atoms, molecules, or other particles) as there are atoms in exactly 12 grams of the carbon-12 isotope. The unit for mole used in calculations is mol. A mole is a unit like a pair, dozen, or gross. A mole is the amount of substance that contains 6.022  1023 particles. Avogadro’s number (6.022  1023) is the number of atoms in exactly 12 g of the carbon-12 isotope.

3.10 The molar mass of an atom is the mass of one mole, 6.022 × 1023 atoms, of that element. Units are g/mol.

3.11 In one year:

9 365 days 24 h 3600 s 2 particles 17

(6.5 10 people) 4.1 10 particles/yr

1 yr 1 day 1 h 1 person

      

23 17

6.022 10 particles 4.1 10 particles/yr

  

Total time 1.510 yr6

3.12 The thickness of the book in miles would be:

23 16

0.0036 in 1 ft 1 mi

(6.022 10 pages) = 3.4 10 mi

1 page 12 in  5280 ft  

The distance, in miles, traveled by light in one year is:

8 12

365 day 24 h 3600 s 3.00 10 m 1 mi

1.00 yr 5.88 10 mi

1 yr 1 day 1 h 1 s 1609 m

       

The thickness of the book in light-years is:

16

12

1 light-yr (3.4 10 mi)

5.88 10 mi

  

5.810 light - yr3

It will take light 5.8  103 years to travel from the first page to the last one!

3.13

6.022 1023S atoms 5.10 mol S

1 mol S

   3.071024S atoms

3.14 9

23

1 mol Co

(6.00 10 Co atoms) =

6.022 10 Co atoms

 

9.961015mol Co

3.15 77.4 g of Ca 1 mol Ca 40.08 g Ca

 1.93 mol Ca

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3.16 Strategy: We are given moles of gold and asked to solve for grams of gold. What conversion factor do we need to convert between moles and grams? Arrange the appropriate conversion factor so moles cancel, and the unit grams is obtained for the answer.

Solution: The conversion factor needed to covert between moles and grams is the molar mass. In the periodic table (see inside front cover of the text), we see that the molar mass of Au is 197.0 g. This can be expressed as

1 mol Au  197.0 g Au

From this equality, we can write two conversion factors.

1 mol Au 197.0 g Au 197.0 g Au and 1 mol Au

The conversion factor on the right is the correct one. Moles will cancel, leaving the unit grams for the answer.

We write

197.0 g Au

= 15.3 mol Au =

1 mol Au

3

? g Au 3.0110 g Au

Check: Does a mass of 3010 g for 15.3 moles of Au seem reasonable? What is the mass of 1 mole of Au?

3.17 (a)

23

200.6 g Hg 1 mol Hg 1 mol Hg 6.022 10 Hg atoms

 

3.3311022g/Hg atom

(b) 23

20.18 g Ne 1 mol Ne 1 mol Ne 6.022 10 Ne atoms

 

3.3511023g/Ne atom

3.18 (a)

Strategy: We can look up the molar mass of arsenic (As) on the periodic table (74.92 g/mol). We want to find the mass of a single atom of arsenic (unit of g/atom). Therefore, we need to convert from the unit mole in the denominator to the unit atom in the denominator. What conversion factor is needed to convert between moles and atoms? Arrange the appropriate conversion factor so mole in the denominator cancels, and the unit atom is obtained in the denominator.

Solution: The conversion factor needed is Avogadro's number. We have 1 mol  6.022  1023 particles (atoms)

From this equality, we can write two conversion factors.

23 23

1 mol As 6.022 10 As atoms

and 1 mol As

6.022 10 As atoms

The conversion factor on the left is the correct one. Moles will cancel, leaving the unit atoms in the denominator of the answer.

We write

23

74.92 g As 1 mol As 1 mol As 6.022 10 As atoms

  

? g/As atom 1.2441022g/As atom

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(b) Follow same method as part (a).

23

58.69 g Ni 1 mol Ni 1 mol Ni 6.022 10 Ni atoms

  

? g/Ni atom 9.7461023g/Ni atom

Check: Should the mass of a single atom of As or Ni be a very small mass?

3.19 12

23

1 mol Pb 207.2 g Pb 1.00 10 Pb atoms

1 mol Pb 6.022 10 Pb atoms

   

3.441010 g Pb

3.20 Strategy: The question asks for atoms of Cu. We cannot convert directly from grams to atoms of copper.

What unit do we need to convert grams of Cu to in order to convert to atoms? What does Avogadro's number represent?

Solution: To calculate the number of Cu atoms, we first must convert grams of Cu to moles of Cu. We use the molar mass of copper as a conversion factor. Once moles of Cu are obtained, we can use Avogadro's number to convert from moles of copper to atoms of copper.

1 mol Cu  63.55 g Cu The conversion factor needed is

1 mol Cu 63.55 g Cu

Avogadro's number is the key to the second conversion. We have 1 mol  6.022  1023 particles (atoms)

From this equality, we can write two conversion factors.

23 23

1 mol Cu 6.022 10 Cu atoms

and 1 mol Cu

6.022 10 Cu atoms

The conversion factor on the right is the one we need because it has number of Cu atoms in the numerator, which is the unit we want for the answer.

Let's complete the two conversions in one step.

grams of Cu  moles of Cu  number of Cu atoms

1 mol Cu 6.022 1023Cu atoms 3.14 g Cu

63.55 g Cu 1 mol Cu

     22

? atoms of Cu 2.9810 Cu atoms

Check: Should 3.14 g of Cu contain fewer than Avogadro's number of atoms? What mass of Cu would contain Avogadro's number of atoms?

3.21 For hydrogen:

1 mol H 6.022 1023H atoms 1.10 g H

1.008 g H 1 mol H

    6.571023H atoms

For chromium:

1 mol Cr 6.022 1023Cr atoms 14.7 g Cr

52.00 g Cr 1 mol Cr

    1.701023Cr atoms

There are more hydrogen atoms than chromium atoms.

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3.22 22

23

1 mol Pb 207.2 g Pb

2 Pb atoms = 6.881 10 g Pb

1 mol Pb 6.022 10 Pb atoms

  

23 4.003 g He 22

(5.1 10 mol He) = 2.0 10 g He

1 mol He

  

2 atoms of lead have a greater mass than 5.1  1023 mol of helium.

3.23 Using the appropriate atomic masses,

(a) CH4 12.01 amu  4(1.008 amu)  16.04 amu (b) NO2 14.01 amu  2(16.00 amu)  46.01 amu (c) SO3 32.07 amu  3(16.00 amu)  80.07 amu (d) C6H6 6(12.01 amu)  6(1.008 amu)  78.11 amu (e) NaI 22.99 amu  126.9 amu  149.9 amu

(f) K2SO4 2(39.10 amu)  32.07 amu  4(16.00 amu)  174.27 amu (g) Ca3(PO4)2 3(40.08 amu)  2(30.97 amu)  8(16.00 amu)  310.2 amu

3.24 Strategy: How do molar masses of different elements combine to give the molar mass of a compound?

Solution: To calculate the molar mass of a compound, we need to sum all the molar masses of the elements in the molecule. For each element, we multiply its molar mass by the number of moles of that element in one mole of the compound. We find molar masses for the elements in the periodic table (inside front cover of the text).

(a) molar mass Li2CO3  2(6.941 g)  12.01 g  3(16.00 g)  73.89 g (b) molar mass CS2  12.01 g  2(32.07 g)  76.15 g

(c) molar mass CHCl3  12.01 g  1.008 g  3(35.45 g)  119.4 g

(d) molar mass C6H8O6  6(12.01 g)  8(1.008 g)  6(16.00 g)  176.12 g (e) molar mass KNO3  39.10 g  14.01 g  3(16.00 g)  101.11 g (f) molar mass Mg3N2  3(24.31 g)  2(14.01 g)  100.95 g

3.25 To find the molar mass (g/mol), we simply divide the mass (in g) by the number of moles.

152 g

0.372 mol  409 g/mol

3.26 The molar mass of acetone, C3H6O, is 58.08 g. We use the molar mass and Avogadro’s number as conversion factors to convert from grams to moles to molecules of acetone.

23

3 6 3 6

3 6

3 6 3 6

1 mol C H O 6.022 10 molecules C H O 0.435 g C H O

58.08 g C H O 1 mol C H O

    4.511021molecules C H O3 6

3.27 We use the molar mass of squaric acid (114.06 g), Avogradro’s number, and the subscripts in the formula of squaric acid, C4H2O4, to convert from grams of squaric acid to moles of squaric acid to molecules of squaric acid, and finally to atoms of C, H, or O. We first convert to molecules of squaric acid.

23 21

4 2 4 4 2 4

4 2 4 4 2 4

4 2 4 4 2 4

1 mol C H O 6.022 10 molecules C H O

1.75 g C H O 9.24 10 molecules C H O

114.06 g C H O 1 mol C H O

    

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Next, we convert to atoms of C, H, and O using the subscripts in the formula as conversion factors.

21

4 2 4

4 2 4

4 atoms C 9.24 10 molecules C H O

1 molecule C H O

   3.701022C atoms

21 4 2 4

4 2 4

2 atoms H 9.24 10 molecules C H O

1 molecule C H O

   1.851022 H atoms

21 4 2 4

4 2 4

4 atoms O 9.24 10 molecules C H O

1 molecule C H O

   3.701022 O atoms

3.28 Strategy: We are asked to solve for the number of N, C, O, and H atoms in 1.68  104 g of urea. We cannot convert directly from grams urea to atoms. What unit do we need to obtain first before we can convert to atoms? How should Avogadro's number be used here? How many atoms of N, C, O, or H are in 1 molecule of urea?

Solution: Let's first calculate the number of N atoms in 1.68  104 g of urea. First, we must convert grams of urea to number of molecules of urea. This calculation is similar to Problem 3.26. The molecular formula of urea shows there are two N atoms in one urea molecule, which will allow us to convert to atoms of N. We need to perform three conversions:

grams of urea  moles of urea  molecules of urea  atoms of N

The conversion factors needed for each step are: 1) the molar mass of urea, 2) Avogadro's number, and 3) the number of N atoms in 1 molecule of urea.

We complete the three conversions in one calculation.

4 1 mol urea 6.022 1023urea molecules 2 N atoms

= (1.68 10 g urea)

60.06 g urea 1 mol urea 1 molecule urea

    

? atoms of N

 3.37  1026 N atoms

The above method utilizes the ratio of molecules (urea) to atoms (nitrogen). We can also solve the problem by reading the formula as the ratio of moles of urea to moles of nitrogen by using the following conversions:

grams of urea  moles of urea  moles of N  atoms of N

Try it.

Check: Does the answer seem reasonable? We have 1.68  104 g urea. How many atoms of N would 60.06 g of urea contain?

We could calculate the number of atoms of the remaining elements in the same manner, or we can use the atom ratios from the molecular formula. The carbon atom to nitrogen atom ratio in a urea molecule is 1:2, the oxygen atom to nitrogen atom ratio is 1:2, and the hydrogen atom to nitrogen atom ration is 4:2.

26 1 C atom

(3.37 10 N atoms)

2 N atoms

    26

? atoms of C 1.6910 C atoms

26 1 O atom

(3.37 10 N atoms)

2 N atoms

    26

? atoms of O 1.6910 O atoms

26 4 H atoms

(3.37 10 N atoms)

2 N atoms

    26

? atoms of H 6.7410 H atoms

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3.29 The molar mass of C19H38O is 282.5 g.

12 1 mol 6.022 1023molecules 1.0 10 g

282.5 g 1 mol

    2.110 molecules9

Notice that even though 1.0  1012 g is an extremely small mass, it still is comprised of over a billion pheromone molecules!

3.30 Mass of water = 2.56 mL 1.00 g = 2.56 g 1.00 mL

Molar mass of H2O  (16.00 g)  2(1.008 g)  18.02 g/mol

2 23 2

2

2 2

1 mol H O 6.022 10 molecules H O

= 2.56 g H O

18.02 g H O 1 mol H O

  

? H O molecules2

 8.56  1022 molecules

3.31 Please see Section 3.4 of the text.

3.32 The relative abundance of each isotope can be determined from the area of the peak in the mass spectrum for that isotope.

3.33 Since there are only two isotopes of carbon, there are only two possibilities for CF4

. (molecular mass 88 amu) and (molecular mass 89 amu)

12 19 13 19

6C 9 4F 6C 9 4F

There would be two peaks in the mass spectrum.

3.34 Since there are two hydrogen isotopes, they can be paired in three ways: 1H-1H, 1H-2H, and 2H-2H. There will then be three choices for each sulfur isotope. We can make a table showing all the possibilities (masses in amu):

32S 33S 34S 36S

1H2 34 35 36 38

1H2H 35 36 37 39

2H2 36 37 38 40

There will be seven peaks of the following mass numbers: 34, 35, 36, 37, 38, 39, and 40.

Very accurate (and expensive!) mass spectrometers can detect the mass difference between two 1H and one

2H. How many peaks would be detected in such a “high resolution” mass spectrum?

3.35 The percent composition is the percent by mass of each element in a compound. For NH3, we would speak of the mass % of nitrogen (N) and the mass % of hydrogen (H) in the compound. What percentage of the mass of a sample of ammonia is due to nitrogen and what percentage of the mass is due to hydrogen?

3.36 If you know the percent composition by mass of an unknown compound, you can determine its empirical formula.

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3.37 An empirical formula tells us which elements are present and the simplest whole-number ratio of their atoms.

A definition of empirical is something that is derived from experiment and observation rather than from theory. When chemists analyze an unknown compound, the first step is usually the determination of the compound’s empirical formula. With additional information, it is possible to deduce the molecular formula.

3.38 The approximate molar mass.

3.39 Molar mass of SnO2  (118.7 g)  2(16.00 g)  150.7 g 118.7 g/mol

150.7 g/mol 100%

  

%Sn 78.77%

(2)(16.00 g/mol) 150.7 g/mol 100%

  

%O 21.23%

3.40 Strategy: Recall the procedure for calculating a percentage. Assume that we have 1 mole of CHCl3. The percent by mass of each element (C, H, and Cl) is given by the mass of that element in 1 mole of CHCl3 divided by the molar mass of CHCl3, then multiplied by 100 to convert from a fractional number to a percentage.

Solution: The molar mass of CHCl3  12.01 g/mol  1.008 g/mol  3(35.45 g/mol)  119.4 g/mol. The percent by mass of each of the elements in CHCl3 is calculated as follows:

12.01 g/mol

%C 100%

119.4 g/mol

   10.06%

1.008 g/mol

%H 100%

119.4 g/mol

   0.8442%

3(35.45) g/mol

%Cl 100%

119.4 g/mol

   89.07%

Check: Do the percentages add to 100%? The sum of the percentages is (10.06%  0.8442%  89.07%)  99.97%. The small discrepancy from 100% is due to the way we rounded off.

3.41 The molar mass of cinnamic alcohol, C9H10O, is 134.17 g/mol.

(a) (9)(12.01 g/mol) 134.17 g/mol 100%

  

%C 80.56%

(10)(1.008 g/mol) 134.17 g/mol 100%

  

%H 7.513%

16.00 g/mol 134.17 g/mol 100%

  

%O 11.93%

(b)

23

9 10 9 10

9 10

9 10 9 10

1 mol C H O 6.022 10 molecules C H O 0.469 g C H O

134.17 g C H O 1 mol C H O

  

 2.11  1021 molecules C9H10O

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3.42 Compound Molar mass (g) N% by mass

(a) (NH2)2CO 60.06 2(14.01 g) 100% = 46.65%

60.06 g 

(b) NH4NO3 80.05 2(14.01 g) 100% = 35.00%

80.05 g 

(c) HNC(NH2)2 59.08 3(14.01 g) 100% = 71.14%

59.08 g 

(d) NH3 17.03 14.01 g 100% = 82.27%

17.03 g Ammonia, NH3, is the richest source of nitrogen on a mass percentage basis.

3.43 Assume you have exactly 100 g of substance.

C

1 mol C

44.4 g C 3.70 mol C

12.01 g C

  

n

H

1 mol H

6.21 g H 6.16 mol H

1.008 g H

  

n

S

1 mol S

39.5 g S 1.23 mol S

32.07 g S

  

n

O

1 mol O

9.86 g O 0.616 mol O

16.00 g O

  

n

Thus, we arrive at the formula C3.70H6.16S1.23O0.616. Dividing by the smallest number of moles (0.616 mole) gives the empirical formula, C6H10S2O.

To determine the molecular formula, divide the molar mass by the empirical mass.

molar mass 162 g empirical molar mass  162.3 g  1

Hence, the molecular formula and the empirical formula are the same, C6H10S2O.

3.44 METHOD 1:

Step 1: Assume you have exactly 100 g of substance. 100 g is a convenient amount, because all the percentages sum to 100%. The percentage of oxygen is found by difference:

100%  (19.8%  2.50%  11.6%)  66.1%

In 100 g of PAN there will be 19.8 g C, 2.50 g H, 11.6 g N, and 66.1 g O.

Step 2: Calculate the number of moles of each element in the compound. Remember, an empirical formula tells us which elements are present and the simplest whole-number ratio of their atoms. This ratio is also a mole ratio. Use the molar masses of these elements as conversion factors to convert to moles.

1 mol C

= 19.8 g C = 1.65 mol C 12.01 g C

Cn

1 mol H

= 2.50 g H = 2.48 mol H 1.008 g H

Hn

1 mol N

= 11.6 g N = 0.828 mol N 14.01 g N

Nn

(10)

1 mol O

= 66.1 g O = 4.13 mol O 16.00 g O

On

Step 3: Try to convert to whole numbers by dividing all the subscripts by the smallest subscript. The formula is C1.65H2.48N0.828O4.13. Dividing the subscripts by 0.828 gives the empirical formula, C2H3NO5.

To determine the molecular formula, remember that the molar mass/empirical mass will be an integer greater than or equal to one.

molar mass

1 (integer values) empirical molar mass 

In this case,

molar mass 120 g empirical molar mass  121.05 g  1

Hence, the molecular formula and the empirical formula are the same, C2H3NO5.

METHOD 2:

Step 1: Multiply the mass % (converted to a decimal) of each element by the molar mass to convert to grams of each element. Then, use the molar mass to convert to moles of each element.

1 mol C

(0.198) (120 g) 1.98 mol C 12.01 g C

    

C 2 mol C

n

1 mol H

(0.0250) (120 g) 2.98 mol H 1.008 g H

    

H 3 mol H

n

1 mol N

(0.116) (120 g) 0.994 mol N 14.01 g N

    

N 1 mol N

n

1 mol O

(0.661) (120 g) 4.96 mol O 16.00 g O

    

O 5 mol O

n

Step 2: Since we used the molar mass to calculate the moles of each element present in the compound, this method directly gives the molecular formula. The formula is C2H3NO5.

Step 3: Try to reduce the molecular formula to a simpler whole number ratio to determine the empirical formula. The formula is already in its simplest whole number ratio. The molecular and empirical formulas are the same. The empirical formula is C2H3NO5.

3.45 2 3 2 3

2 3 2 3

1 mol Fe O 2 mol Fe 24.6 g Fe O

159.7 g Fe O 1 mol Fe O

   0.308 mol Fe

3.46 Using unit factors we convert:

g of Hg  mol Hg  mol S  g S

1 mol Hg 1 mol S 32.07 g S 246 g Hg

200.6 g Hg 1 mol Hg 1 mol S

    

? g S 39.3 g S

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3.47 The balanced equation is: 2Al(s)  3I2(s)  2AlI3(s)

Using unit factors, we convert: g of Al  mol of Al  mol of I2  g of I2

2 2

2

3 mol I 253.8 g I 1 mol Al

20.4 g Al

26.98 g Al 2 mol Al 1 mol I

    288 g I2

3.48 Strategy: Tin(II) fluoride is composed of Sn and F. The mass due to F is based on its percentage by mass in the compound. How do we calculate mass percent of an element?

Solution: First, we must find the mass % of fluorine in SnF2. Then, we convert this percentage to a fraction and multiply by the mass of the compound (24.6 g), to find the mass of fluorine in 24.6 g of SnF2.

The percent by mass of fluorine in tin(II) fluoride, is calculated as follows:

2 2

mass of F in 1 mol SnF

mass % F 100%

molar mass of SnF

 

2(19.00 g)

100% = 24.25% F 156.7 g

 

Converting this percentage to a fraction, we obtain 24.25/100  0.2425.

Next, multiply the fraction by the total mass of the compound.

? g F in 24.6 g SnF2  (0.2425)(24.6 g)  5.97 g F

Check: As a ball-park estimate, note that the mass percent of F is roughly 25 percent, so that a quarter of the mass should be F. One quarter of approximately 24 g is 6 g, which is close to the answer.

Note: This problem could have been worked in a manner similar to Problem 3.46. You could complete the following conversions:

g of SnF2  mol of SnF2  mol of F  g of F

3.49 In each case, assume 100 g of compound.

(a) 2.1 g H 1 mol H 2.1 mol H 1.008 g H

 

1 mol O

65.3 g O 4.08 mol O

16.00 g O

 

1 mol S

32.6 g S 1.02 mol S

32.07 g S

 

This gives the formula H2.1S1.02O4.08. Dividing by 1.02 gives the empirical formula, H2SO4.

(b) 20.2 g Al 1 mol Al 0.749 mol Al 26.98 g Al

 

1 mol Cl

79.8 g Cl 2.25 mol Cl

35.45 g Cl

 

This gives the formula, Al0.749Cl2.25. Dividing by 0.749 gives the empirical formula, AlCl3.

(12)

3.50 (a)

Strategy: In a chemical formula, the subscripts represent the ratio of the number of moles of each element that combine to form the compound. Therefore, we need to convert from mass percent to moles in order to determine the empirical formula. If we assume an exactly 100 g sample of the compound, do we know the mass of each element in the compound? How do we then convert from grams to moles?

Solution: If we have 100 g of the compound, then each percentage can be converted directly to grams. In this sample, there will be 40.1 g of C, 6.6 g of H, and 53.3 g of O. Because the subscripts in the formula represent a mole ratio, we need to convert the grams of each element to moles. The conversion factor needed is the molar mass of each element. Let n represent the number of moles of each element so that

C

1 mol C

40.1 g C 3.34 mol C

12.01 g C

  

n

H

1 mol H

6.6 g H 6.5 mol H

1.008 g H

  

n

O

1 mol O

53.3 g O 3.33 mol O

16.00 g O

  

n

Thus, we arrive at the formula C3.34H6.5O3.33, which gives the identity and the mole ratios of atoms present.

However, chemical formulas are written with whole numbers. Try to convert to whole numbers by dividing all the subscripts by the smallest subscript (3.33).

3.34 1 3.33 

C : 6.5

3.33  2

H : 3.33

3.33  1 O :

This gives the empirical formula, CH2O.

Check: Are the subscripts in CH2O reduced to the smallest whole numbers?

(b) Following the same procedure as part (a), we find:

C

1 mol C

18.4 g C 1.53 mol C

12.01 g C

  

n

N

1 mol N

21.5 g N 1.53 mol N

14.01 g N

  

n

K

1 mol K

60.1 g K 1.54 mol K

39.10 g K

  

n

Dividing by the smallest number of moles (1.53 mol) gives the empirical formula, KCN.

3.51 The molar mass of CaSiO3 is 116.17 g/mol.

40.08 g

%Ca 100%

116.17 g

   34.50%

28.09 g

%Si 100%

116.17 g

   24.18%

(3)(16.00 g)

%O 100%

116.17 g

   41.32%

Check to see that the percentages sum to 100%. (34.50%  24.18%  41.32%)  100.00%

(13)

3.52 The empirical molar mass of CH is approximately 13.02 g. Let's compare this to the molar mass to determine the molecular formula.

Recall that the molar mass divided by the empirical mass will be an integer greater than or equal to one.

molar mass

1 (integer values) empirical molar mass 

In this case,

molar mass 78 g empirical molar mass  13.02 g  6

Thus, there are six CH units in each molecule of the compound, so the molecular formula is (CH)6, or C6H6.

3.53 Find the molar mass corresponding to each formula.

For C4H5N2O: 4(12.01 g)  5(1.008 g)  2(14.01 g)  (16.00 g)  97.10 g For C8H10N4O2: 8(12.01 g)  10(1.008 g)  4(14.01 g)  2(16.00 g)  194.20 g

The molecular formula is C8H10N4O2. 3.54 METHOD 1:

Step 1: Assume you have exactly 100 g of substance. 100 g is a convenient amount, because all the percentages sum to 100%. In 100 g of MSG there will be 35.51 g C, 4.77 g H, 37.85 g O, 8.29 g N, and 13.60 g Na.

Step 2: Calculate the number of moles of each element in the compound. Remember, an empirical formula tells us which elements are present and the simplest whole-number ratio of their atoms. This ratio is also a mole ratio. Let nC, nH, nO, nN, and nNa be the number of moles of elements present. Use the molar masses of these elements as conversion factors to convert to moles.

C

1 mol C

35.51 g C 2.957 mol C

12.01 g C

  

n

H

1 mol H

4.77 g H 4.73 mol H

1.008 g H

  

n

O

1 mol O

37.85 g O 2.366 mol O

16.00 g O

  

n

N

1 mol N

8.29 g N 0.592 mol N

14.01 g N

  

n

Na

1 mol Na

13.60 g Na 0.5916 mol Na

22.99 g Na

  

n

Thus, we arrive at the formula C2.957H4.73O2.366N0.592Na0.5916, which gives the identity and the ratios of atoms present. However, chemical formulas are written with whole numbers.

Step 3: Try to convert to whole numbers by dividing all the subscripts by the smallest subscript.

2.957

= 4.998 5

0.5916 

C : 4.73

= 8.00 0.5916

H : 2.366

= 3.999 4

0.5916 

O : 0.592

= 1.00 0.5916

N : 0.5916

0.5916 = 1 Na :

This gives us the empirical formula for MSG, C5H8O4NNa.

(14)

To determine the molecular formula, remember that the molar mass/empirical mass will be an integer greater than or equal to one.

molar mass

1 (integer values) empirical molar mass 

In this case,

molar mass 169 g empirical molar mass  169.11 g  1

Hence, the molecular formula and the empirical formula are the same, C5H8O4NNa. It should come as no surprise that the empirical and molecular formulas are the same since MSG stands for monosodiumglutamate.

METHOD 2:

Step 1: Multiply the mass % (converted to a decimal) of each element by the molar mass to convert to grams of each element. Then, use the molar mass to convert to moles of each element.

C

1 mol C

(0.3551) (169 g) 5.00 mol C 12.01 g C

   

n

H

1 mol H

(0.0477) (169 g) 8.00 mol H 1.008 g H

   

n

O

1 mol O

(0.3785) (169 g) 4.00 mol O 16.00 g O

   

n

N

1 mol N

(0.0829) (169 g) 1.00 mol N 14.01 g N

   

n

Na

1 mol Na

(0.1360) (169 g) 1.00 mol Na 22.99 g Na

   

n

Step 2: Since we used the molar mass to calculate the moles of each element present in the compound, this method directly gives the molecular formula. The formula is C5H8O4NNa.

3.55 A chemical reaction is a process in which a substance (or substances) is changed into one or more new substances. In this reaction, hydrogen and oxygen, the reactants, are changed into the product, water.

3.56 A chemical equation uses chemical symbols to show what happens during a chemical reaction. A chemical reaction is a process in which a substance (or substances) is changed into one or more new substances.

3.57 To accurately show what happens during a chemical reaction, a chemical equation must be balanced. The Law of Conservation of Mass is obeyed by a balanced chemical equation.

3.58 (g), (l), (s), (aq).

3.59 The balanced equations are as follows:

(a) 2C  O2  2CO (f) 2O3  3O2

(b) 2CO  O2  2CO2 (g) 2H2O2  2H2O  O2 (c) H2  Br2  2HBr (h) N2  3H2  2NH3

(d) 2K  2H2O  2KOH  H2 (i) Zn  2AgCl  ZnCl2  2Ag (e) 2Mg  O2  2MgO (j) S8  8O2  8SO2

(15)

(k) 2NaOH  H2SO4  Na2SO4  2H2O (m) 3KOH  H3PO4  K3PO4  3H2O (l) Cl2  2NaI  2NaCl  I2 (n) CH4  4Br2  CBr4  4HBr 3.60 The balanced equations are as follows:

(a) 2N2O5  2N2O4  O2 (b) 2KNO3  2KNO2  O2 (c) NH4NO3  N2O  2H2O (d) NH4NO2  N2  2H2O (e) 2NaHCO3  Na2CO3  H2O  CO2 (f) P4O10  6H2O  4H3PO4

(g) 2HCl  CaCO3  CaCl2  H2O  CO2 (h) 2Al  3H2SO4  Al2(SO4)3  3H2 (i) CO2  2KOH  K2CO3  H2O (j) CH4  2O2  CO2  2H2O

(k) Be2C  4H2O  2Be(OH)2  CH4 (l) 3Cu  8HNO3  3Cu(NO3)2  2NO  4H2O (m) S  6HNO3  H2SO4  6NO2  2H2O (n) 2NH3  3CuO  3Cu  N2  3H2O

3.61 Stoichiometry is the quantitative study of reactants and products in a chemical reaction; therefore, it is based on the Law of Conservation of Mass. A balanced chemical equation is essential to solving stoichiometric problems so that the “mole method” can be applied correctly.

3.62 The steps of the mole method are shown in Figure 3.8 of the text.

3.63 On the reactants side there are 8 A atoms and 4 B atoms. On the products side, there are 4 C atoms and 4 D atoms. Writing an equation,

8A  4B  4C  4D

Chemical equations are typically written with the smallest set of whole number coefficients. Dividing the equation by four gives,

2A  B  C  D

The correct answer is choice (c).

3.64 On the reactants side there are 6 A atoms and 4 B atoms. On the products side, there are 4 C atoms and 2 D atoms. Writing an equation,

6A  4B  4C  2D

Chemical equations are typically written with the smallest set of whole number coefficients. Dividing the equation by two gives,

3A  2B  2C  D

The correct answer is choice (d).

3.65 The mole ratio from the balanced equation is 2 moles CO2 : 2 moles CO.

2 mol CO2

3.60 mol CO

2 mol CO

  3.60 mol CO2

(16)

3.66 Si(s)  2Cl2(g)   SiCl4(l)

Strategy: Looking at the balanced equation, how do we compare the amounts of Cl2 and SiCl4? We can compare them based on the mole ratio from the balanced equation.

Solution: Because the balanced equation is given in the problem, the mole ratio between Cl2 and SiCl4 is known: 2 moles Cl2  1 mole SiCl4. From this relationship, we have two conversion factors.

2 4

4 2

2 mol Cl 1 mol SiCl 1 mol SiCl and 2 mol Cl

Which conversion factor is needed to convert from moles of SiCl4 to moles of Cl2? The conversion factor on the left is the correct one. Moles of SiCl4 will cancel, leaving units of "mol Cl2" for the answer. We

calculate moles of Cl2 reacted as follows:

4 2

4

2 mol Cl 0.507 mol SiCl

1 mol SiCl

  

2 2

? mol Cl reacted 1.01 mol Cl

Check: Does the answer seem reasonable? Should the moles of Cl2 reacted be double the moles of SiCl4

produced?

3.67 Starting with the amount of ammonia produced (6.0 moles), we can use the mole ratio from the balanced equation to calculate the moles of H2 and N2 that reacted to produce 6.0 moles of NH3.

3H2(g)  N2(g)  2NH3(g)

2 3 2

3

3 mol H

? mol H 6.0 mol NH

2 mol NH

   9.0 mol H2

2 3 2

3

1 mol N

? mol N 6.0 mol NH

2 mol NH

   3.0 mol N2

3.68 Starting with the 5.0 moles of C4H10, we can use the mole ratio from the balanced equation to calculate the moles of CO2 formed.

2C4H10(g)  13O2(g)  8CO2(g)  10H2O(l)

2 4 10 2 2

4 10

8 mol CO

? mol CO 5.0 mol C H 20 mol CO

2 mol C H

    2.010 mol CO1 2

3.69 It is convenient to use the unit ton-mol in this problem. We normally use a g-mol. 1 g-mol SO2 has a mass of 64.07 g. In a similar manner, 1 ton-mol of SO2 has a mass of 64.07 tons. We need to complete the following conversions: tons SO2  ton-mol SO2  ton-mol S  ton S.

7 2

2

2 2

1 ton-mol SO 1 ton-mol S 32.07 ton S (2.6 10 tons SO )

64.07 ton SO 1 ton-mol SO 1 ton-mol S

     1.310 tons S7

3.70 (a) 2NaHCO3   Na2CO3  H2O  CO2

(17)

(b) Molar mass NaHCO3  22.99 g  1.008 g  12.01 g  3(16.00 g)  84.01 g Molar mass CO2  12.01 g  2(16.00 g)  44.01 g

The balanced equation shows one mole of CO2 formed from two moles of NaHCO3.

3 3

2 2

2 2 3

2 mol NaHCO 84.01 g NaHCO 1 mol CO

= 20.5 g CO

44.01 g CO 1 mol CO 1 mol NaHCO

  

mass NaHCO3

 78.3 g NaHCO3

3.71 The balanced equation shows a mole ratio of 1 mole HCN : 1 mole KCN.

1 mol KCN 1 mol HCN 27.03 g HCN 0.140 g KCN

65.12 g KCN 1 mol KCN 1 mol HCN

    0.0581 g HCN

3.72 C6H12O6   2C2H5OH  2CO2 glucose ethanol

Strategy: We compare glucose and ethanol based on the mole ratio in the balanced equation. Before we can determine moles of ethanol produced, we need to convert to moles of glucose. What conversion factor is needed to convert from grams of glucose to moles of glucose? Once moles of ethanol are obtained, another conversion factor is needed to convert from moles of ethanol to grams of ethanol.

Solution: The molar mass of glucose will allow us to convert from grams of glucose to moles of glucose.

The molar mass of glucose  6(12.01 g)  12(1.008 g)  6(16.00 g)  180.16 g. The balanced equation is given, so the mole ratio between glucose and ethanol is known; that is 1 mole glucose  2 moles ethanol.

Finally, the molar mass of ethanol will convert moles of ethanol to grams of ethanol. This sequence of three conversions is summarized as follows:

grams of glucose  moles of glucose  moles of ethanol  grams of ethanol

6 12 6 2 5 2 5

6 12 6

6 12 6 6 12 6 2 5

1 mol C H O 2 mol C H OH 46.07 g C H OH 500.4 g C H O

180.16 g C H O 1 mol C H O 1 mol C H OH

   

2 5

? g C H OH

 255.9 g C2H5OH

Check: Does the answer seem reasonable? Should the mass of ethanol produced be approximately half the mass of glucose reacted? Twice as many moles of ethanol are produced compared to the moles of glucose reacted, but the molar mass of ethanol is about one-fourth that of glucose.

The liters of ethanol can be calculated from the density and the mass of ethanol.

volume mass

density

255.9 g

= = 324 mL =

0.789 g/mL

Volume of ethanol obtained 0.324 L

3.73 The mass of water lost is just the difference between the initial and final masses.

Mass H2O lost  15.01 g  9.60 g  5.41 g

2 2 2

2

1 mol H O moles of H O 5.41 g H O

18.02 g H O

   0.300 mol H O2

(18)

3.74 The balanced equation shows that eight moles of KCN are needed to combine with four moles of Au.

1 mol Au 8 mol KCN

29.0 g Au =

197.0 g Au 4 mol Au

  

? mol KCN 0.294 mol KCN

3.75 The balanced equation is: CaCO3(s)  CaO(s)  CO2(g)

3 3

3 3

1 mol CaCO

1000 g 1 mol CaO 56.08 g CaO

1.0 kg CaCO

1 kg 100.09 g CaCO 1 mol CaCO 1 mol CaO

     5.610 g CaO2

3.76 (a) NH4NO3(s)   N2O(g)  2H2O(g)

(b) Starting with moles of NH4NO3, we can use the mole ratio from the balanced equation to find moles of N2O. Once we have moles of N2O, we can use the molar mass of N2O to convert to grams of N2O.

Combining the two conversions into one calculation, we have:

mol NH4NO3  mol N2O  g N2O

2 2

4 3

4 3 2

1 mol N O 44.02 g N O 0.46 mol NH NO

1 mol NH NO 1 mol N O

    1

2 2

? g N O 2.010 g N O

3.77 The quantity of ammonia needed is:

8 4 2 4 3 3

4 2 4

4 2 4 4 2 4 3

2 mol NH 17.03 g NH

1 mol (NH ) SO 1 kg

1.00 10 g (NH ) SO

132.15 g (NH ) SO 1 mol (NH ) SO 1 mol NH 1000 g

    

 2.58  104 kg NH3

3.78 The balanced equation for the decomposition is : 2KClO3(s)   2KCl(s)  3O2(g)

3 2 2

3

3 3 2

1 mol KClO 3 mol O 32.00 g O 46.0 g KClO

122.55 g KClO 2 mol KClO 1 mol O

    

2 2

? g O 18.0 g O

3.79 The reactant used up first in a reaction is called the limiting reagent. Excess reagents are the reactants present in quantities greater than necessary to react with the quantity of the limiting reagent. The maximum amount of product formed depends on how much of the limiting reagent is present. When this reactant is used up, no more product can be formed. In a reaction with one reactant, the one reactant is by definition the limiting reagent.

3.80 If you are making sandwiches and have four pieces of bread, you can only make two sandwiches, no matter how much mayonnaise, mustard, sandwich meat, etc. that you have. The bread limits the number of sandwiches that you can make.

3.81 2A  B  C

(a) The number of B atoms shown in the diagram is 5. The balanced equation shows 2 moles A  1 mole B.

Therefore, we need 10 atoms of A to react completely with 5 atoms of B. There are only 8 atoms of A present in the diagram. There are not enough atoms of A to react completely with B.

A is the limiting reagent.

(19)

(b) There are 8 atoms of A. Since the mole ratio between A and B is 2:1, 4 atoms of B will react with 8 atoms of A, leaving 1 atom of B in excess. The mole ratio between A and C is also 2:1. When 8 atoms of A react, 4 molecules of C will be produced.

3.82 N2  3H2  2NH3

9 moles of H2 will react with 3 moles of N2, leaving 1 mole of H2 in excess. The mole ratio between N2 and NH3 is 1:2. When 3 moles of N2 react, 6 moles of NH3 will be produced.

3.83 This is a limiting reagent problem. Let's calculate the moles of NO2 produced assuming complete reaction for each reactant.

2NO(g)  O2(g)  2NO2(g)

2 2

2 mol NO

0.886 mol NO 0.886 mol NO

2 mol NO

 

2 2 2

2

2 mol NO

0.503 mol O 1.01 mol NO

1 mol O

 

NO is the limiting reagent; it limits the amount of product produced. The amount of product produced is 0.886 mole NO2.

3.84 Strategy: Note that this reaction gives the amounts of both reactants, so it is likely to be a limiting reagent problem. The reactant that produces fewer moles of product is the limiting reagent because it limits the amount of product that can be produced. How do we convert from the amount of reactant to amount of product? Perform this calculation for each reactant, then compare the moles of product, NO2, formed by the given amounts of O3 and NO to determine which reactant is the limiting reagent.

Solution: We carry out two separate calculations. First, starting with 0.740 g O3, we calculate the number of moles of NO2 that could be produced if all the O3 reacted. We complete the following conversions.

grams of O3  moles of O3  moles of NO2

H2

NH3

B C

(20)

Combining these two conversions into one calculation, we write

3 2

2 3 2

3 3

1 mol O 1 mol NO

? mol NO 0.740 g O 0 0154 mol NO

48 00 g O 1 mol O

    .

.

Second, starting with 0.670 g of NO, we complete similar conversions.

grams of NO  moles of NO  moles of NO2 Combining these two conversions into one calculation, we write

2 2 2

1 mol NO 1 mol NO

? mol NO 0.670 g NO 0 0223 mol NO

30 01 g NO 1 mol NO

    .

.

The initial amount of O3 limits the amount of product that can be formed; therefore, it is the limiting reagent.

The problem asks for grams of NO2 produced. We already know the moles of NO2 produced, 0.0154 mole.

Use the molar mass of NO2 as a conversion factor to convert to grams (Molar mass NO2  46.01 g).

2 2

2

46.01 g NO 0.0154 mol NO

1 mol NO

  

2 2

? g NO 0.709 g NO

Check: Does your answer seem reasonable? 0.0154 mole of product is formed. What is the mass of 1 mole of NO2?

Strategy: Working backwards, we can determine the amount of NO that reacted to produce 0.0154 mole of NO2. The amount of NO left over is the difference between the initial amount and the amount reacted.

Solution: Starting with 0.0154 mole of NO2, we can determine the moles of NO that reacted using the mole ratio from the balanced equation. We can calculate the initial moles of NO starting with 0.670 g and using molar mass of NO as a conversion factor.

2

2

1 mol NO

mol NO reacted 0.0154 mol NO 0.0154 mol NO 1 mol NO

  

1 mol NO

mol NO initial 0.670 g NO 0.0223 mol NO 30 01 g NO

  

.

mol NO remaining  mol NO initial  mol NO reacted.

mol NO remaining  0.0223 mol NO  0.0154 mol NO  0.0069 mol NO

3.85 (a) The balanced equation is: C3H8(g)  5O2(g)  3CO2(g)  4H2O(l)

(b) The balanced equation shows a mole ratio of 3 moles CO2 : 1 mole C3H8. The mass of CO2 produced is:

2 2

3 8

3 8 2

3 mol CO 44.01 g CO 3.65 mol C H

1 mol C H 1 mol CO

   482 g CO2

3.86 This is a limiting reagent problem. Let's calculate the moles of Cl2 produced assuming complete reaction for each reactant.

2 2 2

2

1 mol Cl

0.86 mol MnO = 0.86 mol Cl 1 mol MnO

(21)

2 2

1 mol Cl 1 mol HCl

48.2 g HCl = 0.330 mol Cl

36.46 g HCl 4 mol HCl

 

HCl is the limiting reagent; it limits the amount of product produced. It will be used up first. The amount of product produced is 0.330 mole Cl2. Let's convert this to grams.

2 2

2

70.90 g Cl

0.330 mol Cl =

1 mol Cl

 

2 2

? g Cl 23.4 g Cl

3.87 The theoretical yield of a reaction is the amount of product that would result if all the limiting reagent reacted. When the limiting reactant is used up, no more product can be formed.

3.88 There are many reasons why the actual yield is less than the theoretical yield. Some reactions are reversible, so they do not proceed 100% from reactants to product.. There could be impurities in the starting materials.

Sometimes it is difficult to recover all the product. There may be side reactions that lead to additional products.

3.89 The balanced equation is given: CaF2  H2SO4  CaSO4  2HF

The balanced equation shows a mole ratio of 2 moles HF : 1 mole CaF2. The theoretical yield of HF is:

3 2

2

2 2

1 mol CaF 2 mol HF 20.01 g HF 1 kg

(6.00 10 g CaF ) 3.08 kg HF

78.08 g CaF 1 mol CaF 1 mol HF 1000 g

     

The actual yield is given in the problem (2.86 kg HF).

actual yield

% yield 100%

theoretical yield

 

2.86 kg 3.08 kg 100%

  

% yield 92.9%

3.90 (a) Start with a balanced chemical equation. It’s given in the problem. We use NG as an abbreviation for nitroglycerin. The molar mass of NG  227.1 g/mol.

4C3H5N3O9   6N2  12CO2  10H2O  O2 Map out the following strategy to solve this problem.

g NG  mol NG  mol O2  g O2 Calculate the grams of O2 using the strategy above.

2 2 2

2

1 mol O 32.00 g O 1 mol NG

2.00 10 g NG

227.1 g NG 4 mol NG 1 mol O

     

2 2

? g O 7.05 g O

(b) The theoretical yield was calculated in part (a), and the actual yield is given in the problem (6.55 g).

The percent yield is:

actual yield

% yield 100%

theoretical yield

 

2 2

6.55 g O

100% = 7.05 g O

 

% yield 92.9%

(22)

3.91 The balanced equation shows a mole ratio of 1 mole TiO2 : 1 mole FeTiO3. The molar mass of FeTiO3 is 151.73 g/mol, and the molar mass of TiO2 is 79.88 g/mol. The theoretical yield of TiO2 is:

6 3 2 2

3

3 3 2

1 mol FeTiO 1 mol TiO 79.88 g TiO 1 kg 8.00 10 g FeTiO

151.73 g FeTiO 1 mol FeTiO 1 mol TiO 1000 g

    

 4.21  103 kg TiO2

The actual yield is given in the problem (3.67  103 kg TiO2).

3 3

actual yield 3.67 10 kg

% yield 100% 100%

theoretical yield 4.21 10 kg

     

87.2%

3.92 This is a limiting reagent problem. Let's calculate the moles of Li3N produced assuming complete reaction for each reactant.

6Li(s)  N2(g)  2Li3N(s)

3

3

2 mol Li N 1 mol Li

12.3 g Li 0.591 mol Li N

6.941 g Li 6 mol Li

  

3

2 2 3

2 2

2 mol Li N 1 mol N

33.6 g N 2.40 mol Li N

28.02 g N 1 mol N

  

Li is the limiting reagent; it limits the amount of product produced. The amount of product produced is 0.591 mole Li3N. Let's convert this to grams.

3 3 3

3

34.83 g Li N

? g Li N 0.591 mol Li N

1 mol Li N

   20.6 g Li N3

This is the theoretical yield of Li3N. The actual yield is given in the problem (5.89 g). The percent yield is:

actual yield 5.89 g

% yield 100% 100%

theoretical yield 20.6 g

     28.6%

3.93 All the carbon from the hydrocarbon reactant ends up in CO2, and all the hydrogen from the hydrocarbon reactant ends up in water. In the diagram, we find 4 CO2 molecules and 6 H2O molecules. This gives a ratio between carbon and hydrogen of 4:12. We write the formula C4H12, which reduces to the empirical formula CH3. The empirical molar mass equals approximately 15 g, which is half the molar mass of the hydrocarbon.

Thus, the molecular formula is double the empirical formula or C2H6. Since this is a combustion reaction, the other reactant is O2. We write:

C2H6  O2  CO2  H2O Balancing the equation,

2C2H6  7O2  4CO2  6H2O

3.94 2H2(g)  O2(g)  2H2O(g)

We start with 8 molecules of H2 and 3 molecules of O2. The balanced equation shows 2 moles H2  1 mole O2. If 3 molecules of O2 react, 6 molecules of H2 will react, leaving 2 molecules of H2 in excess. The balanced equation also shows 1 mole O2  2 moles H2O. If 3 molecules of O2 react, 6 molecules of H2O will be produced.

(23)

After complete reaction, there will be 2 molecules of H2 and 6 molecules of H2O. The correct diagram is choice (b).

3.95 First, let's convert to moles of HNO3 produced.

4

3 3 3

3

1 mol HNO 2000 lb 453.6 g

1.00 ton HNO 1.44 10 mol HNO

1 ton 1 1b 63.02 g HNO

    

Now, we will work in the reverse direction to calculate the amount of reactant needed to produce 1.44  103 mol of HNO3. Realize that since the problem says to assume an 80% yield for each step, the amount of reactant needed in each step will be larger by a factor of 100%

80% , compared to a standard stoichiometry calculation where a 100% yield is assumed.

Referring to the balanced equation in the last step, we calculate the moles of NO2.

4 2 4

3 2

3

2 mol NO 100%

(1.44 10 mol HNO ) 3.60 10 mol NO

1 mol HNO 80%

    

Now, let's calculate the amount of NO needed to produce 3.60  104 mol NO2. Following the same

procedure as above, and referring to the balanced equation in the middle step, we calculate the moles of NO.

4 4

2

2

1 mol NO 100%

(3.60 10 mol NO ) 4.50 10 mol NO

1 mol NO 80%

    

Now, let's calculate the amount of NH3 needed to produce 4.5  104 mol NO. Referring to the balanced equation in the first step, the moles of NH3 is:

4 3 4

3

4 mol NH 100%

(4.50 10 mol NO) 5.63 10 mol NH

4 mol NO 80%

    

Finally, converting to grams of NH3:

4 3

3

3

17.03 g NH 5.63 10 mol NH

1 mol NH

   9.5910 g NH5 3

3.96 We assume that all the Cl in the compound ends up as HCl and all the O ends up as H2O. Therefore, we need to find the number of moles of Cl in HCl and the number of moles of O in H2O.

1 mol HCl 1 mol Cl

mol Cl 0.233 g HCl 0.00639 mol Cl

36.46 g HCl 1 mol HCl

   

2 2

2 2

1 mol H O 1 mol O

mol O 0.403 g H O 0.0224 mol O

18.02 g H O 1 mol H O

   

Dividing by the smallest number of moles (0.00639 mole) gives the formula, ClO3.5. Multiplying both subscripts by two gives the empirical formula, Cl2O7.

3.97 The number of moles of Y in 84.10 g of Y is:

1 mol X 1 mol Y

27.22 g X 0.8145 mol Y

33.42 g X 1 mol X

  

Figure

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