**Problem #7:** 0.158 g of a barium halide is completely precipitated with H_{2}SO_{4} and 0.124 g of BaSO_{4} is collected. What is the formula of the barium halide?

**Solution #1**

(1) (137.3 / 233.4) x 0.124 = 0.0728 g of Ba in sample (137.3 is the atomic weight of Ba; 233.4 is the molecular weight of BaSO_{4})

(2) (0.0728 g / 0.158 g) x 100 = 46.1% barium

(3) the halides are F, Cl, Br, I. a brute force solution of each formula in turn yields BaBr_{2} as the only one with 46% barium.

**Solution #2**

(1) The reaction equation is:

BaX_{2} + H_{2}SO_{4} ---> BaSO_{4} + 2HX

(2) 0.124 g ÷ 233.4 g/mol = 0.0005314 mole of BaSO_{4}

(3) This required 0.0005314 mole of BaX_{2} since there is a one-to-one ratio between BaX_{2} and BaSO_{4}

(4) Therefore 0.158 g ÷ (137.3 + 2x) g/mol = 0.0005314 mol

(5) x = 80 therefore BaBr_{2}

**Problem #8:** A sample of cocaine, C_{17}H_{21}O_{4}N, is diluted with sugar, C_{12}H_{22}O_{11}. When a 1.00 mg sample of this mixture is burned, 2.00 mg CO_{2} is formed. What is the percent cocaine in the mixture?

**Solution #1**

(1) The solution requires two simultaneous equations in two unknowns. Let x be the amount of cocaine and y be the amount of sugar, both amounts in the sample of 1.00 mg.

First equation

x mg + y mg = 1.00 mg sample

Second equation

0.673x + 0.421y = 0.5458

0.673x equals the milligrams of CARBON in cocaine, 0.421y equals the milligrams of CARBON in sugar and 0.5458 equals the milligrams in 2.00 mg of CO_{2}. (0.673 is the decimal percentage carbon in cocaine and 0.421 is the decimal percentage carbon in sugar.) That 2.00 mg of CO_{2} is produced from the 1.00 mg sample is given in the problem.

(2) Proceed to solve by setting x = 1 minus y (rearrange the first equation. The use it to eliminate the x in the second equation:

0.673(1 - y) + 0.421y = 0.5458

(3) After some simple algebra, we get:

0.252y = 0.1272

Therefore y = 50.5% (the mass percent of sugar in the sample) and cocaine is 49.5%

**Solution #2**

This is a wrong solution. See if you can spot the error.

(1) 2.00 mg x (12.011 / 44.0098) = 0.546 mg of carbon in the CO_{2}.

(2) 0.546 mg x (17/29) = 0.320 mg carbon from the cocaine in the sample.

(3a) Cocaine is 67% carbon, so 0.67x = 0.320; x = 47.8% (This is cocaine's mass percent of the sample.)

Did you see the mistake? No? Before discussing the mistake, lets do the same calculation in #3a, but for the sugar.

(3b) Sugar is 42% carbon, so 0.42x = 0.226. The 0.226 came from 0.546 minus 0.320. x = 53.8%

The problem is that 43.8 + 53.8 does not equal 100, it equals 101.8. If this was the right calculation technique, step 3b would yield an answer that added up to 100%. It did not.

The problem is in the ratio of 17/29. It implies a one-to-one molar ratio of cocaine and sugar molecules and, while that might be the case, it is probably not the case. The 17/29 ratio says that there are two molecules and one contributes 17 carbons and the other 12. That is only true in a one-to-one ratio.

In order to determine the true molar ratio of cocaine to sugar, you must FIRST know the mass distribution. Only then can you calculate the moles of cocaine and sugar and then look at the ratio of moles between cocaine and sugar. On a molar basis, the cocaine to sugar ratio is 1.05 to 0.95, which is a 52.5% to 47.5% ratio. The ChemTeam will allow you to check the answers yourself.