Question
Asked Sep 26, 2019
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A sample of a substance (containing only C, H, and N) is burned in oxygen.
3.939 g of CO21.209 g of H2O and 4.028 g of NO are the sole products of combustion.
What is the empirical formula of the compound?

 

What was the mass of the initial sample burned?

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Expert Answer

Step 1

Given that the mass of CO2, H2O and NO is 3.939 g , 1.209 g and 4.028 g respectively.

The mass of Carbon, Hydrogen and Nitrogen can be calculated using the simple mole concept.

12 gC1.074gC
1mol CO2
MassofC 3.939 gx molCO,
44g
Massof H 1.209 g xmol H2O
18g
2g H
1molH2O
=0.134g H
14gC1.879gN
Imol NO
Massof N 4.028gx.
1mol NO
30g
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12 gC1.074gC 1mol CO2 MassofC 3.939 gx molCO, 44g Massof H 1.209 g xmol H2O 18g 2g H 1molH2O =0.134g H 14gC1.879gN Imol NO Massof N 4.028gx. 1mol NO 30g

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Step 2

Number of moles of Carbon, Hydrogen and Oxygen can be calculated by the equation given below:

Mass
No.of molesof C=.
Atomic mass
1.074g0,0895 mol
12g/mol
No.of moles of H=948-0.1340mol
1g/mo
No.of moles of N1.1g0.1342mol
14g/mol
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Mass No.of molesof C=. Atomic mass 1.074g0,0895 mol 12g/mol No.of moles of H=948-0.1340mol 1g/mo No.of moles of N1.1g0.1342mol 14g/mol

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Step 3

In-order to get the empirical formula, first the number of moles has to be divided with the lowest...

Dividing by the lowest integer,0.0895
0.1340
Н:
0.0895
0.0895
С:
0.0895
0.1342
-=1.5
-=1.5 ; N:
=1
0.0895
In-order to convert to whole number, muliply with 2
C:1x2=2 H:1.5x2-3 ; C:1.5x2 3
Emperical formula =C,H,N
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Dividing by the lowest integer,0.0895 0.1340 Н: 0.0895 0.0895 С: 0.0895 0.1342 -=1.5 -=1.5 ; N: =1 0.0895 In-order to convert to whole number, muliply with 2 C:1x2=2 H:1.5x2-3 ; C:1.5x2 3 Emperical formula =C,H,N

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