Chapter 3 - Stoichiometry

Question 1

3.2

Write the equation for a Combination reaction.

Answer 1

A + B —-> C

Question 2

3.2

Write the equation for a Decomposition reaction.

Answer 2

C —–> A + B

Question 3

3.2

Write the equation for a Combustion reaction.

Answer 3

Fuel + O2(g) —> CO2(g) + H2O(l)

Question 4

3.3

Define “formula weight”.

Answer 4

The sum of the atomic weights of each atom in its chemical formula.

Ex. H2SO4 = 2(1.0 amu) + 32.1 amu + 4(16.0 amu) = 98.1 amu

Question 5

3.3

Define “molecular weight”.

Answer 5

The mass of the collection of atoms represented by the chemical formula for a molecule.

Ex. C6H12O6 = 6(12.0 amu) + 12(1.0 amu) + 6(16.0 amu) = 180.0 amu

Question 6

3.3

Write the equation for % element in a substance.

Answer 6

% element= ((# atoms of said element * atomic weight of element) / formula weight of compound) x 100

Question 7

3.3 Samp. Ex

Calculate the % of C, H, & O (by mass) in C12H22O11.

Answer 7

% C = ((12 * 12.01 amu)/342.22 amu) x 100 = 42.1%
% H = ((22 * 1.01 amu)/342.22 amu) x 100 = 6.4%
% O = ((11 * 16 amu)/342.22 amu) x 100 = 51.5%

Question 8

3.4

1 mol NO3- ions = x ions

Answer 8

6.022 x 10^23

Question 9

3.4 Samp. Ex.

Without using a calculator, arrange the following samples in order of increasing numbers of C atoms: 12g C-12, 1 mol C2H2, 9 x 10^23 molecules of CO2.

Answer 9

12g C-12 (6 x 10^23 atoms) < 9 x 10^23 molecules of CO2 (9 x 10^23 atoms because each molecule has 1 C atom) < 1 mol C2H2 (~12 x 10^23 atoms)

Question 10

3.4 Practice Ex.

Arrange the following samples in order of increasing # of O atoms: 1 mol H2O, 1 mol CO2, 3 x 10^23 molecules O3

Answer 10

1 mol H2O (6 x 10^23 atoms) < 3 x 10^23 molecules O3 (9 x 10^23 atoms) < 1 mol CO2 (12 x 10^23 atoms)

Question 11

3.4

Define “molar mass”.

Answer 11

The mass in grams of 1 mol of a substance.

The molar mass (in g/mol) of any substance is always numerically equal to its formula weight (in amu).

Question 12

3.5 Samp. Ex.

Ascorbic acid (vitamin C) contains 40.92% C, 4.58% H, and 54.50% O by mass. What is its empirical formula?

Answer 12

40. 92g (1 mol/12.01g) = 3.407 mol C
41. 58g (1 mol/1.01g) = 4.54 mol H
42. 50g (1 mol/16g) = 3.406 mol O

Divide them all by 3.407, the smallest number.

C = 1, H = 1.33, O = 1

Multiply by 3 to get whole numbers.

C= 3, H = 4, O = 3

C3H4O3

Question 13

3.5 Prac. Ex.

A 5.325g sample of methyl benzoate contains 3.758g C, 0.316 g H, and 1.251g O. What is its empirical formula?

Answer 13

3. 758g C (1 mol / 12.01g) = .3129 mol C
4. 316g H (1 mol / 1.01g) = .313 mol H
5. 251g O (1 mol / 16g) = .0782 mol O

Divide by .0782.

C: 4, H: 4, O: 1

C4H4O

Question 14

3.5

Whole-number multiple = ? / ?

Answer 14

Molecular weight/empirical formula weight

Question 15

3.5 Samp. Ex

Mesitylene has an empirical formula of C3H4. The molecular weight is 121 amu. What’s the molecular formula?

Answer 15

121/40.07 = 3

Multiply by 3.

C9H12

Question 16

3.5 Samp. Ex.

Isopropyl alcohol is composed of C, H, and O. Combustion of 0.255g of isopropyl alcohol produces .561g of CO2 & .306g of H2O. Determine isopropyl alcohol’s empirical formula.

Answer 16

.561g CO2 (1 mol / 44.01g)(1 mol C/ 1 mol CO2)(12.01g C / 1 mol C) = 0.153g C

.306g H2O (1 mol / 18.02g)(2 mol H / 1 mol H2O)(1.01g H / 1 mol H) = 0.0343g H

Mass of O = .255g isopropyl alcohol - 0.153g C - 0.0343g H = .0677g O

Gram to mol conversions.

0.0128 mol C, 0.0340 mol H, 0.0043 mol O.

Divide by .0043 to produce 3:8:1 ratio.

C3H8O

Question 17

3.6 Give It Some Thought (p. 99)

When 1.57 mol O2 reacts with H2 to form H2O, how many moles of H2 are consumed in the process?

Answer 17

2 H2 + O2 –> 2 H2O

1.57 mol O2 (2 mol H2 / 1 mol O2) = 3.14 mol H2

Question 18

3.7

How do you find the limiting reactant?

Answer 18

Figure out which reactant produces the smallest amount of product. That’s the limiter.

ex.
150g H2 (1 mol / 2.02g)(2 mol H2O / 1 mol H2)(18.02g H2O / 1 mol H2) = 74g H2O

1500g O2 (1 mol / 32g)(2 mol H2O / 1 mol O2)(18.02g H2O / 1 mol) = 1300g H2O

Hydrogen limits.

Question 19

3.7

Define “theoretical yield”.

Answer 19

The quantity of product that’s calculated to form when all of the limiting reactant reacts.

Question 20

3.7

Define “actual yield”.

Answer 20

The amount of product actually obtained in a eraction.

Question 21

3.7

Percent yield = (? / ?) x 100

Find the percent yield of adipic acid. The theoretical yield is 43.5g. The actual yield is 33.5g.

Answer 21

(Actual / theoretical) x 100

(33.5g / 43.5g) x 100 = 77.0%

Question 22

3.1 - Give It Some Thought (p. 80)

How many atoms of Mg, O, and H are represented by 3 Mg(OH)2?

Answer 22

3 atoms Mg, 6 atoms O, 6 atoms H