### Bonus Empirical Formula Problem Answers 3 & 4

Problem #3: For the reaction represented by the equation:

CX4 + 2 O2 ---> CO2 + 2 X2O

9.0 g of CX4 completely reacts with 1.74 g of oxygen. What is the approximate molar mass of X?

Solution #1

(1) moles O2 = 1.74 g ÷ 32.0 g/mol = 0.0544 mol.

(2) The CX4 to O2 ratio is 1 to 2. Therefore, 1 is to 2 as x is to 0.0544. x = 0.0272 mol of CX4 reacted.

(3) 9.0 g ÷ 0.0272 mol = 331 g/mol. This is the molecular weight of CX4

(4) Subtract 12 from 331 to get 319, which is the weight contribution of X4.

(5) Dividing by four gets 79.8. The nearest atomic weight on the periodic table is that of bromine. The formula is CBr4.

Here are some some pieces of alternate solutions that use bits of the above solution and then go off in a "different" direction. In each case, solve for x.

(A) 9.0 g ÷ (12.011 + 4x) g/mol = 0.0272 mol
(B) (1.74 g / 32.0 g/mol) ÷ (9.0 g / [12 + 4x] g/mol) = 2 ÷ 1

Solution #2

I would have NEVER thought of this:

(1) 64.0 /1.74 = 36.8 (I think the proper unit here is mol¯1)

(2) 36.8 mol¯1 x 9.0 g = 331 g/mol

Pretty wild, huh? The two steps could be put together this way: 64.0 ÷ 1.74 = x ÷ 9. By the way, the 64.0 comes from the 2O2.

Solution #3

(1) 1.74 g is the 2O2 in the equation. Half of it goes to the CO2. This is 0.87 g of oxygen.

(2) In CO2, there is 72.7% oxygen (32 ÷ 44), so (0.727) (x) = 0.87. x = 1.196 g, the total mass of CO2 which contains 0.87 g of oxygen.

(3) Carbon is 27.27% carbon (12 ÷ 44) in CO2, so (0.2727) (1.196) = 0.326 g of carbon in CO2.

(4) 0.326 g of carbon is 0.02717 mol of carbon. This is also the mol of CX4 present.

(5) 9.0 minus 0.326 equals 8.674g, the mass of X4 in CX4. 8.674 ÷ 4 = 2.1685 g, the mass of one X in the 9.0 g sample.

(6) 2.1685 g ÷ 0.02717 mole = 79.8 g/mol, the atomic weight of bromine.

Problem #4: A mixture of NaCl and NaBr weighing 1.234 g is heated with chlorine gas, which converts the mixture completely to NaCl. The total mass of NaCl is now 1.129 g. What are the mass percentages of NaCl and NaBr in the original sample?

Solution

(1) The NaCl/NaBr mix weighed 1.234 g and the pure NaCl after conversion is 1.129 g. The difference (0.105 g) is the amount of bromine (NOT NaBr) in the original sample.

(2) 0.105 g ÷ 80.1 g/mol = 0.00131 mol of Br

(3) Therefore, 0.00131 mole of NaBr in sample. Justification is in the equation:

2NaBr + Cl2 ---> 2NaCl + Br2

Think of the Br2 as 2 Br.

(4) (0.00131 mol) (103.1 g/mol) = 0.135 g of NaBr in original mix

(5) 1.234 g - 0.135 g = 1.099 g of NaCl in original mix

(6) 1.099 / 1.234 = 89% NaCl, NaBr = 11%