# How to compute molecular formula?

I have the following question to solve:

Tungsten, $\ce{W}$, and chlorine, $\ce{Cl}$, form a series of compounds with the following compositions: \begin{array}{rr} \text{Mass % W} & \text{Mass % Cl}\\ \hline 72.17 & 27.83\\ 56.45 & 43.55\\ 50.91 & 49.09\\ 46.36 & 53.64\\ \end{array} If a molecule of each compound contains only one tungsten atom, what are the formulas for the four compounds?

For one gram of tungsten, chlorine has mass \begin{align} \frac{27.83}{72.17} &= 0.3302~\mathrm{g}\\[3pt] \frac{43.55}{56.45} &= 0.7715~\mathrm{g}\\[3pt] \frac{49.09}{50.91} &= 0.9643~\mathrm{g}\\[3pt] \frac{53.64}{46.36} &= 1.157~\mathrm{g}.\\ \end{align}

Since $\frac{0.7715}{0.3302}\approx\frac{7}{3}$, $\frac{0.9643}{0.3302}\approx3$, and $\frac{1.157}{0.3302}\approx\frac{7}{2}$, the number of atoms of chlorine for a given mass of tungsten are respectively in the ratio $6:14:18:21$.

So if a molecule of each compound contains only one tungsten atom, the formulas are $\ce{WCl_{6}}$, $\ce{WCl_{14}}$, $\ce{WCl_{18}}$, and $\ce{WCl_{21}}$.

Is this correct?

• No. Look up "empirical formula" in a chemistry book or Google it. May 13, 2012 at 15:49
• This question served as a bad example for the homework policy. (It might be quite a good fit nowadays though.) Feb 23, 2016 at 8:46

We don't have a homework policy yet, but I'm quite sure that it will be against providing full answers.

Hints:

• You're supposed to take into account molar masses. The ratios in a formula are molar ratios, so you need to use those, not mass ratios.

• Read up a bit on emperical formulae, as Janice suggested.

• This answer is preventing the question from being deleted no? Jun 10, 2015 at 21:30
• Just my opinion, the question should be kept since the homework policy uses it as an example of a bad question. That's especially true since the other two bad examples there have already been deleted. Sep 13, 2015 at 19:49
• It seems that we may have become more lax on homework questions because if OP posted this question today, I think this would be considered pretty good. Sep 30, 2015 at 18:25