Interesting problem. Thanks for posting.
C2H2 + (3/2)02 ====> H2O + 2CO2
CH4 + 2O2 =====> 2H2O + CO2
The molar mass of C2H2 = 2*12 + 2*1 = 26
The molar mass of CH4 = 1*12 + 4*1 = 16
The number of moles of C2H2 = x
The number of moles of CH4 = y
26x + 16y = 230.9 grams
For water we get (from the C2H2). Water has a molar mass of 2*1 + 16 = 18
x*18 See the balanced equation to see what it is the same number of moles as C2H2
From the methane we get
y*18
2*y* 18. Again see the balanced equation to see where that 2 came from.
18x + 36y is the total amount of water.
Now for the CO2. CO2 has a molar mass of 12 + 2*16 = 44
From C2H2 we get 2*44*x = 88x grams of CO2
From CH4 we get 1*y*44 grams of CO2
88x + 44y for CO2
Now we total to get the grand total of water and CO2
18x + 44y + 88x + 44y = 972.7 grams total.
106x + 88y = 972.7
Two equations, two unknowns, we should be able to solve this problem
26x + 16y = 230.9
106x + 88y = 972.7
I'm not going to go through the math unless you request me to do so.
x = 8.03 moles
y = 1.38 moles
The initial amount of C2H2 was 8.03 * 26 = 208.78
The initial amount of CH4 was 16*1.38 = 22.08
The total (as a check is 230.86 which is pretty close to the given amount.
So Methane's mass in the initial givens was 22.08 grams.
The empirical formula gives the relative ratio of atoms in each element. Therefore, it simplifies the whole numbers. For example C2H6 can be reduced to CH3 because they share the greatest common factor (2). The answer to this question would be #1. C4H10 can be reduced to C2H5 because 4 and 10 are both divisible by 2. C2H5 can not be reduced any further. All of the other options do not have a greatest common factor making them a empirical formula.
The answer is 2.7gm/cm^3 . Density is mass divided by volume .
