Fairly easy Mass % problem. Don't need the answer, just the explanation?

Question

Here it is:
When Carbon-containing compounds are burned in a limited amount of air, some CO as well as some CO2 is produced. A gaseous product mixture is 35.0 mass % CO and 65.0 mass % CO2. What is mass % of C in the mixture?

Now, for the mass of CO2 I got 46g and for CO I got 30g. Now, I have the formula, but whenever I use it I seem to get a weird answer. The formula was:
(Moles of X in formula x Molar Mass of X (g/mol)/mass (g) of 1 mol of compound) x 100

I plugged in 3 x 16/76g x 100 and got somewhere around 63%, but where does the 65 and 35% tie in!?!

If you can sort this out I would be much obliged. Thanks guys.

Merry Christmas!

Answer 1

Since the mass percentages of the compounds present in the mixture are given and the mass percentage of one of the elements is asked, you don't need the masses of the compounds. In fact it is impossible to calculate the exact amounts of the compounds by using the given data.

Assume you have 100 grams of the gaseous mixture.
Mass of CO = 35 g
Mass of CO2 = 65 g

Molar mass of CO = 28 g/mol, in which mass of C is 12 g.
Molar mass of CO2 = 44 g/ mol, in which mass of C is 12 g.

Mass of C in 35 g of CO = 35 (12/28) = 15.0 g
Mass of C in 65 g of CO2 = 65 (12/44) = 17.7 g

Total mass of C in 100 g mixture : 15.0 + 17.7 = 32.7 g
Since this mass of C is in 100 g mixture, it also represents the mass percentage of C = 32.7 %.

Check the answer, by calculating the mass percent of O.
Mass of O in 35 g of CO = 35 (16/28) = 20.0 g
Mass of O in 65 g of CO2 = 65 (32/44) = 47.3 g
Total mass of O in 100 g mixture : 20.0 + 47.3 = 67.3 g which corresponds to 67.3%
C = 32.7 %
O = 67.3 %
Total = 100 %

Answer 2

In 100g you have 65% CO2 and 35% CO
In the 65g of CO2 xg =C
In the 35g ofCO xg=C
In 1 mol CO2= 32g O, 12g C; 44g in all
12/44=% =>0.27% multiplied by 65g=17.55g C in CO2
12/28=%=>0.43 % multiplied by 35g=15g C in CO
You add them together and =32.6g or 32.6 % C in the mixture
I hope this makes sense.

Answer 3

Atomic weights: C=12 O=16 CO2=44 CO=28

Let the mixture be called M.

35gCO/100gM x 1molCO/28gCO x 1molC/1molCO x 12gC/1molC = 0.15gC/1gM

65gCO2/100gM x 1molCO2/44gCO2 x 1molC/2molO x 12gC/1mol = 0.089gC/1g M

0.15gC + 0.09gC = 0.24gC/1gM x 100% = 24%

Merry Christmas to you also,