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An unknown compound contains only C , H , and O . Combustion of 4.90 g of this compound produced 11.1 g CO2 and 4.56 g H2O . What is the empirical formula of the unknown compound? Insert subscripts as needed.

Answer :

cristisp93

empirical formula C₄H₈O

Explanation:

First we need to calculate the mass of carbon from the carbon dioxide.

if in       44 g of carbon dioxide there are 12 g of carbon

then in  11.1 g of carbon dioxide there are X g of carbon

X = (11.1 × 12) / 44 = 3.03 g carbon

if in       18 g of water there are 2 g of hydrogen

then in  4.56 g of water there are Y g of hydrogen

Y = (4.56 × 2) / 18 = 0.507 g hydrogen

mass the unknown compound = mass of carbon + mass of hydrogen + mass of oxygen

mass of oxygen = mass the unknown compound - (mass of carbon + mass of hydrogen)

mass of oxygen = 4.90 - (3.03 + 0.507) = 1.02 g oxygen

To find the empirical formula of the unknown compound we divide the mass of each element my its atomic wight:

for carbon  3.03 / 12 = 0.253

for hydrogen  0.507 / 1 = 0.507

for oxygen  1.02 / 16 = 0.0637

The numbers we get are divided to the smallest one which is 0.0637:

for carbon 0.253 / 0.0637 ≈ 4

for hydrogen 0.507 / 0.0637 ≈ 8

for oxygen 0.0637 / 0.0637 = 1

The empirical formula of the unknown compound is:

C₄H₈O

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