<em>Answer:</em>
- The molecular formula will be C6H10.
<em>Explanation;</em>
<em>Given Data:</em>
- mass of compound = 0.8300 g
- mass of H2O = 0.9102 g
- mass of CO2 = 2.668 g
Solution:
% of C = (mass of CO2/mass of compound)× (12/44)×100
% of C = (2.66÷0.8300)×(12/44)× 100 = 87.40 %
% of H = (mass of H2O/mass of compound)× (2/18)×100
% of H = (0.910÷0.8300)× (2/18)×100 = 12.18 %
Niiw we calculate moles of C and H from % as follow
moles of C = 87.40/12 = 7.2 mol
moles of H = 12.18/1 = 12.18 mol
In we take atomic ratio
C : H
12.18/7.2 : 7.2/7.2
1.69 : 1
multiply atomic ratio by 3 to get whole no.
C : H = 3 (1.69:1) = 5 : 3
So empirical formula is C3H5.
<em>Calculations of molecular formula:</em>
empirical formula mass of C3H5 = 41
n = molar mass of compound /empirical formula mass
n = 82.1/41 = 2
Molecular mass = n × empirical formula = 2 × C3H5 = C6H10