Calculate the molar mass of the following:
(i) \(\text{H}_2\text{O}\) (ii) \(\text{CO}_2\) (iii) \(\text{CH}_4\)
(i) \(\text{H}_2\text{O}\) (ii) \(\text{CO}_2\) (iii) \(\text{CH}_4\)
Solution and Explanation
(i) \(\text{H}_2\text{O}\):
The molecular mass of water, \(\text{H}_2\text{O}\)
\(= (2 \times \text{Atomic mass of hydrogen}) + (1 \times \text{Atomic mass of oxygen})\)
= [2(1.0084) + 1(16.00 u)]
= 18.016
= 18.02 u
(ii) \(\text{CO}_2\):
The molecular mass of carbon dioxide, \(\text{CO}_2\)
\(= (1 \times \text{Atomic mass of carbon}) + (2 \times \text{Atomic mass of oxygen})\)
= [1(12.011 u) + 2 (16.00 u)]
= 12.011 u + 32.00 u
= 44.01 u
(iii) \(\text{CH}_4\):
The molecular mass of methane, \(\text{CH}_4\)
\(= (1 \times \text{Atomic mass of carbon}) + (4 \times \text{Atomic mass of hydrogen})\)
= [1(12.011 u) + 4 (1.008 u)]
= 12.011 u + 4.032 u
= 16.043 u
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Concepts Used:
Molecular Mass of Polymers
It is described as the distribution rather than a specific number due to the occurrence of polymerization in such a way as to produce different chain lengths. Polymer MW is derived as follows:
\[M_{W} = \sum^{N}_{i=1} w_{i}MW_{i}.\]Where,
wi = the weight fraction of polymer chains having a molecular weight of MWi.
The MW is typically measured by light dispersing experiments. The degree of dispersing arises from the molecule size and, thus, molecular weight dispensation can be mathematically set on the total scattering created by the sample.