Chemistry Help Please! How do I find % composition?

Question

A white powder, consisting of a simple mixture of tartaric acid (C4H6O6) and citric acid (C6H8O7) was analysed to determine the elemental composition. Combustion of a 433.7-mg sample produced 533.6 mg of CO2 and 158.0 mg of H2O.

Then I have to find the %C, H, O, etc. by mass. Can someone please explain this to me? Its my last question and I cant get it :(

Answer 1

due to conservation of mass...
we know that 533.6+158-433.7= 257.9 mg oxygen from OUTSIDE sources were used

next write the combustion equations
2C4H6O6 + 5O2 -> 8CO2 + 6H2O
(mm = 150, C only 48, = 32%)
and

2C6H8O7 + 9O2 -> 12CO2 + 8H2O
(mm = 192, C only = 72, = 37.5%)
now...

the C in all of the CO2 formed comes from the acids and no where else... C accounts for 27.273% of the mass of CO2

thus from 533.6 mg of CO2, we know that 145.53 mg is carbon

we know that the mixture is 433.7 mg, thus
C accounts for 145.53/433.7 * 100% = 33.555%
this is expected since the % should lie between the two values of the acids, which are 32% for tartaric and 37.5% for citric

thus we let the mixture consist of x mg of tartaric acid and y mg of citric acid...
so obviously
x + y = 433.7
and also
(32%x + 37.5%y)/ (x+y) = 33.555%
then multiply by 100 on both sides and substitute x+y with 433.7
32x + 37.5 y = (33.555) (433.7) = 14552.7273
then since x + y = 433.7, multiply with 32
32x + 32y = 13878.4
thus
5.5 y = 14552.73 - 13878.4
y = 122.6054545
then x = 311.0945455

thus the mixture consists of 311.1 mg tartaric acid and 122.6 mg citric acid.

the steps for finding % H and O are now straighforward... do the rest of your homework urself :p

Answer 2

even however your question is poorly written, and that i actual want to verbally abuse you for looking forward to people to help you once you in all probability do not even understand what you probably did write, i'm going to anticipate i understand what the difficulty needs you to do. you are able to desire to discover the full molar mass of the compound, then destroy it down into the guy factors. Take the mass of the element in the compound over the full molar mass, and additionally you get the % composition of that element. to illustrate, in the 1st one, you will discover the full molar mass. i'm going to anticipate you're able to try this, as i'm not likely to waste my time doing it for you. Then, enable's say you want to discover the % composition of the Mg, you will possibly take the molar mass of Mg on my own, divided by utilising the full molar mass of compound.