An unknown compound has the formula $C_xH_yO_z$. You burn 0.1523 g of the compound and isolate 0.3718 g of $CO_2$ and 0.1522 g of $H_2O$. What is the empirical formula of the compound?

Md Ariful Islam

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First, work out the mass of each element that was present in the original compound. Carbon is always present as $CO_2$ in the ratio (12.011 g / 44.0098 g), and hydrogen is always present as $H_2O$ in the ratio (2.0158 g / 18.0152 g).

So you need to work out the mass of carbon in 0.3718 g of $CO_2$ and the mass of hydrogen in 0.1522 g of water..

Carbon: 0.3718 x (12.011 / 44.0098) = 0.10147 g
Hydrogen: 0.1522 x (2.0158 / 18.0152) = 0.01703 g

You can determine the mass of oxygen by difference. 0.1523 - 0.10147 - 0.01703 = 0.0338 g
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Next, convert each of these to numbers of moles:

Carbon: 0.10147 / 12.011 = 0.00845 mol
Hydrogen: 0.01703 / 1.0079 = 0.01689 mol
Oxygen: 0.0338 / 15.994 = 0.002113 mol
Remember that you're working out moles of hydrogen atoms here, not moles of molecular hydrogen gas. So use the atomic weight of hydrogen 1.0079.
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Next, divide each number of moles by the lowest value, so that you get as close as possible to whole numbers.

Carbon: 0.00845 / 0.002113 = 4
Hydrogen: 0.001689 / 0.002113 = 8
Oxygen: 0.002113/ 0.002113 = 1

So the emprical formula is $C_4H_8O$

The molar mass is 72.1 g/mol - work out the mass of $C_4H_8O$ - this is (12 x 4) + (8 x 1) + 16 = 72. Therefore the molecular formula is the same as the empirical..

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