# A compound containing only carbon and hydrogen has a carbon-to-hydrogen mass ratio of 11.89. Which carbon-to-hydrogen mass ratio is possible for another compound composed only of carbon and hydrogen?

## a) 2.50 b) 3.97 c) 4.66 d) 7.89 I'm asking for a friend, so a thorough explanation would be extremely helpful!

Sep 27, 2016

Trying to present a possible Answer

We know that the molar mass of carbon and hydrogen are

$C = 12 \text{g/mol" and H = 1"g/mol}$

When Carbon Hydrogen mass ratio is 11.89

The ratio of no. Of atoms of C and H is given by

$\frac{C}{H} = \frac{\frac{11.89}{12}}{\frac{1}{1}} \approx \frac{1}{1}$

So empirical formula $C H \to \text{possible}$

Now from the list

a)
When Carbon Hydrogen mass ratio is 2.5

The ratio of no. Of atoms of C and H is given by

$\frac{C}{H} = \frac{\frac{2.5}{12}}{\frac{1}{1}} \approx \frac{5}{24}$

So empirical formula ${C}_{5} {H}_{24} \to \text{ not possible}$
Beceause the molecular formula of alkane with 5 C-atoms in the molecule is ${C}_{5} {H}_{12}$. So the maximum number of hydrogen with 5 C should not exceed 12.

b)
When Carbon Hydrogen mass ratio is 3.97

The ratio of no. Of atoms of C and H is given by

$\frac{C}{H} = \frac{\frac{3.97}{12}}{\frac{1}{1}} \approx \frac{1}{3}$

So empirical formula $C {H}_{3} \to \text{possible}$ This is found in ethane having MF ${C}_{2} {H}_{6}$

c)
When Carbon Hydrogen mass ratio is 4.66

The ratio of no. Of atoms of C and H is given by

$\frac{C}{H} = \frac{\frac{4.66}{12}}{\frac{1}{1}} \approx \frac{7}{18}$

So empirical formula ${C}_{7} {H}_{18} \to \text{not possible}$ Since maximum number of H with 7C-atoms is $7 \times 2 + 2 = 16$.So it has been exceeded.

d)
When Carbon Hydrogen mass ratio is 7.89

The ratio of no. Of atoms of C and H is given by

$\frac{C}{H} = \frac{\frac{7.89}{12}}{\frac{1}{1}} \approx \frac{2}{3}$

So empirical formula ${C}_{2} {H}_{3} \to \text{ possible}$
Since the molecular formula of butyne or butadiene ${C}_{4} {H}_{6}$ supports the empirical formula
${C}_{2} {H}_{3}$

Jan 4, 2018

This is an example of the Law of Multiple Proportions.

If a carbon-to-hydrogen mass ratio is $11.89$, and then since every carbon/hydrogen atom has the same mass as any other carbon/hydrogen atom (respectively), then some other carbon-and-hydrogen-containing compound must exist that is a whole number multiple of the given ratio.

(Note that it does not mean it is the only mass ratio that can exist. Decimal multiples of the mass ratio are allowed, but do not guarantee a real compound.)

You are given the molar mass ratio ${M}_{C} / {M}_{H} = 11.89$. Well, $\text{1 mol}$ of anything is $\text{1 mol}$ of anything else, so a molar mass ratio is also a mass ratio, i.e.

M_X/M_Y = ("g X/mol")/("g Y/mol") = "g X"/"g Y" -= m_X/m_Y

Therefore, for some ${\text{C"_x"H}}_{y}$:

$\frac{11.89}{2.50} = 4.8$, not a whole number.

$\frac{11.89}{3.97} = 2.99 \approx 3$, so this is possible.

$\frac{11.89}{4.66} = 2.55$, not a whole number.

$\frac{11.89}{7.89} = 1.50$, not a whole number.

Having this ratio, we could identify the empirical formula:

$11.89 = \left(x \cdot \text{12.011 g C/mol")/(y cdot "1.0079 g H/mol}\right)$

If $x = y$, then we have a ratio of $11.92$, i.e. the number of $\text{C}$ atoms is equal to the number of $\text{H}$ atoms. This is reasonably close to the given $11.89$. As a result, we can see that the empirical formula is $\text{CH}$.

If the ratio is $3.97$, we could choose $\left(i\right)$ that $x \to \frac{1}{3} x$ (not reasonable), or $\left(i i\right)$ that $y \to 3 y$.

• Choice $\left(i\right)$ overcomplicates things because it adds a step of multiplying everything by $3$ to not end up with a fractional atom.

• Choice $\left(i i\right)$ gives $\textcolor{b l u e}{{\text{CH}}_{3}}$ as the empirical (and molecular) formula right away:

${m}_{C} / {m}_{H} = \left(1 \cdot \text{12.011 g/mol")/(3 cdot "1.0079 g/mol}\right) = 3.97$

and this correlates with the molecular formula ${\text{C"_2"H}}_{6}$ (ethane).

Note that in general, the mass ratio is not equal to the mol ratio.