Question

Combustion of 30.42 g of a compound containing only carbon, hydrogen, and oxygen produces 35.21 g CO2 and 14.42 g H2O. What is the empirical formula of the compound

Answer 1

Answer:

C2H4O3

Explanation:

We would have to do some preparations between before solving it the normal way. The main goal is to get the masses of the Individual elements. So here goes;

We can get the mass of C from CO2 using the following steps:

1 mole of CO2 has a mass of 44g (Molar mass) and contains 12g of C.

How did we know the molar mass of CO2 is 44g?

Easy. 1 mole of C = 12, 1 mole of O = 16

But we have two O’s so the total mass of O = (2 * 6) = 32

Total mass of CO2 = mass of C + Mass of O = 12 + 32 = 44

So if 44g of CO2 contains 12g of C, how much of C would be present in 35.21g CO2.

12 = 44

X = 35.21

X = (35.21 * 12) / 44 = 9.603g

We can also get the mass of H from H2O. 1 mole of H2O has a mass of 18g and contains 2g of H.

How did we know the molar mass of H2O is 18g?

Easy. 1 mole of H = 1, 1 mole of O = 16

But we have two H’s so the total mass of H = (2 * 1) = 2

Total mass of H2O = mass of H + Mass of O = 2 + 16 = 18

So how much of H would be present in 14.42g of H2O?

2 = 18

X =14.42

X = (14.42 * 2 ) / 18 = 1.602g

Now we have the masses of C and H. But the question says the compound contains the C, H and O.

So we still have to calculate the mass of Oxygen. We obtain this from;

Mass of Compound = Mass of Carbon + Mass of Oxygen + Mass of Hydrogen

Mass of Oxygen = Mass of compound – (Mass of Carbon + Mass of Hydrogen)

Mass of Oxygen = 30.42 – (9.603 + 1.602)

Mass of Oxygen = 30.42 - 11.205  = 19.215

Now we have all the masses so we are good too go. Let’s have our table.

Elements Carbon (C) Hydrogen (H) Oxygen (O)

Mass        9.603             1.602           19.215

                0.800            1.602           1.2001 (Divide by molar mass)

                1                    2                   1.5       (Divide by lowest number)

                2                    4                   3        (Convert to simple integers by * 2)

The Empirical formula of the compound is C2H4O3