Determine percent of element in sample/compound

An 8.570 mg sample was burned in an oven. The mass of the CO2 collected was 10.20 mg and the mass of the water collected was 1.396 mg.

The Carrius method was used. A 5.19 x 20^-2 gram sample gave 5.40 x 10^-2 of BaSO4.

The chloride ion was precipitated by adding AgNO3. A 3.62 mg sample gave 2.33 x 10^-3 grams of AgCl.

N: 6.33%

Is it:
C=32%
H=11%
S=14%
Cl=16%
O=20%

or:
C=27%
H=11%
S=14%
Cl=25%
O=17%

(percentage values are rounded and therefore do not equal 100)

Question

Determine percent of element in sample/compound

An 8.570 mg sample was burned in an oven. The mass of the CO2 collected was 10.20 mg and the mass of the water collected was 1.396 mg.

The Carrius method was used. A 5.19 x 20^-2 gram sample gave 5.40 x 10^-2 of BaSO4.

The chloride ion was precipitated by adding AgNO3. A 3.62 mg sample gave 2.33 x 10^-3 grams of AgCl.

N: 6.33%

Is it:
C=32%
H=11%
S=14%
Cl=16%
O=20%

or:
C=27%
H=11%
S=14%
Cl=25%
O=17%

(percentage values are rounded and therefore do not equal 100)

jfraser · Answer

I'm not familiar with the names of the methods, but each one produces a particular product that has one of the atoms of the original compound in it.

You find the original mass of carbon in the first test, by using the mass of $CO_2$ produced, and the $mass \space percent$ of the carbon in each gram of $CO_2$, then do the same thing with the hydrogen, and the other two methods, to find the mass of sulfur, then chlorine.

$$0.01020 \cancel{g \space CO_2} * (\frac{12g \space C}{\cancel{44g \space CO_2}}) = 0.002782g \space C $$Which is 2.782mg

Since the whole sample weighed 8.570mg, you find the percentage of each element out of that total and you've got your answer