### Theory:

Calculate the gram molecular mass of the following compounds.

i. $$H_2O$$
ii. $$CO_2$$
iii. $$(Ca)_3(PO_4)_2$$

 i. $$H_2O$$

Data:

Atomic mass of $$H=1$$

Atomic mass of $$O=16$$

Solution:

$\begin{array}{l}\mathit{Gram}\phantom{\rule{0.147em}{0ex}}\mathit{molecular}\phantom{\rule{0.147em}{0ex}}\mathit{mass}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}{H}_{2}O\\ =\left(2×1\right)+\left(1×16\right)\\ =2+16\\ =18\end{array}$

Hence, the gram molecular mass of $$H_2O=18$$ $$g$$.

 ii. $$CO_2$$

Data:

Atomic mass of $$C=12$$

Atomic mass of $$O=16$$

Solution:

$\begin{array}{l}\mathit{Gram}\phantom{\rule{0.147em}{0ex}}\mathit{molecular}\phantom{\rule{0.147em}{0ex}}\mathit{mass}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}{\mathit{CO}}_{2}\\ =\left(1×12\right)+\left(2×16\right)\\ =12+32\\ =44\end{array}$

Hence, the gram molecular mass of $$CO_2=44$$ $$g$$.

 iii. $$(Ca)_3(PO_4)_2$$

Data:

Atomic mass of $$Ca=40$$

Atomic mass of $$P=30$$

Atomic mass of $$O=16$$

Solution:

$\begin{array}{l}\mathit{Gram}\phantom{\rule{0.147em}{0ex}}\mathit{molecular}\phantom{\rule{0.147em}{0ex}}\mathit{mass}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}{\mathit{Ca}}_{3}{\left({\mathit{PO}}_{4}\right)}_{2}\\ =\left(3×40\right)+\left[30+\left(4×16\right)\right]×2\\ =120+\left[30+64\right]×2\\ =120+\left(94×2\right)\\ =308\end{array}$

Hence, the gram molecular mass of $$(Ca)_3(PO_4)_2=308$$ $$g$$.