# A certain compound contains only C, H, and N. Combustion of 0.125 g of this compound produces 0.172 g of H2O and 0.279 g of CO2. Find the mass percentages of C, H, and N and the empirical formula...

A certain compound contains only C, H, and N. Combustion of 0.125 g of this compound produces 0.172 g of H2O and 0.279 g of CO2.

Find the mass percentages of C, H, and N and the empirical formula of this compound.

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Let us say the empirical formula is C_xH_yN_z

`C_xH_yN_z + rarr CO_2+H_2O+NO_2`

Mole ratio

`C:CO_2 = 1:1`

`H:H_2O = 2:1`

`N:NO_2 = 1:1`

Molar masses in g/mol

C = 12

O = 16

N = 14

`CO_2 = 44`

`H_2O = 18`

Amount of `CO_2` formed `= 0.279/44 = 0.0063`

Amount of `H_2O` formed = 0.172/18 = 0.0096

Amount of C in compound `= 0.0063xx12 = 0.0756g`

Amount of H in compound `= 2xx0.0096xx1 = 0.0192g`

Amount of N in  the compound `= 0.125-0.0756-0.0192 = 0.0302g`

% of C in compound `= 0.0756/0.125xx100% = 60.48%`

% of H in compound `= 0.0192/0.125xx100% = 15.36%`

% of N in compound `= 100-60.48-15.36 = 24.16%`

Amount of N moles in the compound `= 0.0302/14 = 0.0022`

`C_xH_yN_z = 0.0063:0.0196:0.0022 = 3:8:1`

So the emperical formular is `C_3H_8N`