5 Steps to a 5 AP Chemistry 2019

Stoichiometry ‹ 101

❯ Answer and Explanation

(a) The percent oxygen (53.3%) is determined by subtracting the carbon and the hydrogen

from 100%. Assuming there are 100 g of sample gives the grams of each element as

being numerically equivalent to the percent. Dividing the grams by the molar mass of

each element gives the moles of each.

For C: 40.0/12.01 = 3.33 Divide each of

these by the

smallest (3.33)

C = 1

For H: 6.71/1.008 = 6.66 H = 2

For O: 53.3/16.00 = 3.33 O = 1

This gives the empirical formula CH 2 O.

You get 1 point for correctly determining any of the elements and 1 point for get-

ting the complete empirical formula correct, if you show your work.

(b) Using HA to represent the monoprotic acid, the balanced equation for the titration

reaction is

HA(aq) + NaOH(aq) → NaA(aq) + H 2 O(l)

The moles of acid may then be calculated:

(^) ()














45.00 mL NaOH

0.1000 mol NaOH

1000  mL

1 mol HA

1 mol NaOH

= 4.500 × 10 –3 mole HA

The molecular mass is

0.2720 g/4.500 × 10 –3 mol = 60.44 g/mol

You get 1 point for the correct number of moles of HA (or NaOH) and 1 point

for the correct final answer, if you show your work.

(c) There are several methods to solve this problem. One way is to use the ideal gas equa-

tion, as done here. The equation and the value of R are in the exam booklet. First, find

the moles: n = PV/RT. Do not forget: you MUST change temperature to Kelvin.

n = (1.00 atm)(300.0 mL)(1 L/1,000 mL)/(0.0821 L atm/mol K)(373 K)

n = 9.80 × 10 –3 mole

The molecular mass is

1.18 g/9.80 × 10 –3 mol = 120 g/mol

You get 1 point for getting any part of the calculation correct and 1 point for getting

the correct final answer, if you show your work.

(d) The approximate formula mass from the empirical (CH 2 O) formula is

12 + 2(1) + 16 = 30 g/mol

For part b: (60.44 g/mol)/(30 g/mol) = 2

Molecular formula = 2 × Empirical formula = C 2 H 4 O 2

For part c: (120 g/mol)/(30 g/mol) = 4

Molecular formula = 4 × Empirical formula = C 4 H 8 O 4

You get 1 point for each correct molecular formula. If you got the wrong answer

in part a, you can still get credit for one or both molecular formulas if you used the

part a value correctly.