S1.4 Counting particles by mass: The mole

Molar mass of CO₂: $M=12.01+2(16.00)=44.01\approx 44{\text{ g mol}}^{-1}$. Molar mass of H₂O: $M=2(1.01)+16.00=18.02\approx 18{\text{ g mol}}^{-1}$.

$n({\text{CO}}_2)=\frac{132}{44}=3.00\text{ mol}$ $n({\text{H}}_2\text{O})=\frac{36}{18}=2.00\text{ mol}$

Since the number of molecules $N=n\times N_A$, the ratio of molecules equals the ratio of moles: $\frac{N({\text{CO}}_2)}{N({\text{H}}_2\text{O})}=\frac{3.00}{2.00}=\frac{3}{2}$

With $132\text{ g}$ of CO₂: $n=132/44=3.00\text{ mol}$. With $36\text{ g}$ of H₂O: $n=36/18=2.00\text{ mol}$. Ratio = $3:2$. But looking at the options, $\frac{3}{2}$ is option B. Let me recheck: the correct ratio ${\text{CO}}_2:{\text{H}}_2\text{O}=3:2$, which matches option B, $\frac{3}{2}$.

We need the ratio of molecules of CO₂ to H₂O. Since the number of molecules is proportional to moles, we just need the mole ratio.

$n({\text{CO}}_2)=132/44=3.00\text{ mol}$; $n({\text{H}}_2\text{O})=36/18=2.00\text{ mol}$. Ratio = $3:2=\frac{3}{2}$.

$\frac{1}{1}$: This would require equal moles, but $3\ne 2$. $\frac{2}{3}$: This is the inverted ratio (H₂O to CO₂). $\frac{1}{2}$: This would require 1 mol CO₂ for every 2 mol H₂O, which is not the case here.

The ratio of CO₂ molecules to H₂O molecules is $3:2=\frac{3}{2}$, which is option B.

The molar mass $M$ can be found by rearranging $n=\frac{m}{M}$ to give: $M=\frac{m}{n}$

$M=\frac{28.0\text{ g}}{0.400\text{ mol}}=70.0{\text{ g mol}}^{-1}$

The molar mass is $70.0{\text{ g mol}}^{-1}$. This is consistent with the identity of the substance (for reference, germanium has $A_r\approx 72.6$, so this could be a molecular compound).

We need to find the molar mass given both the mass and the number of moles of a sample using $M=m/n$.

This would come from dividing by the wrong factor: $0.400\times 28.0=11.2$. This incorrectly multiplies $m\times n$ rather than dividing, giving a nonsensical unit of $\text{g}\cdot \text{mol}$.

$44.0{\text{ g mol}}^{-1}$ is the molar mass of ${\text{CO}}_2$, but $28.0/0.400=70.0\ne 44.0$. This is a common distractor based on a well-known molar mass.

$112$ would result from $28.0\times 4=112$, incorrectly multiplying the mass by $1/0.4\times$ something. The correct calculation gives $28.0/0.400=70.0$.

$M=28.0/0.400=70.0{\text{ g mol}}^{-1}$, confirming the answer is $70.0{\text{ g mol}}^{-1}$.