Determining the formula of zirconium oxide

Question

I am doing the following question (from the 1988 HSC Chemistry paper):

In the 19th Century the atomic weight of zirconium (Zr) was known but there was considerable doubt about the formula of zirconium oxide. Experiments determined the relative weight of the volatile chloride to be about 233.

Assuming that zirconium in the oxide has the same valency, this evidence suggests that the formula for the oxide is:

(a) $\ce{ZrO}$

(b) $\ce{Zr2O3}$

(c) $\ce{ZrO2}$

(d) $\ce{ZrO4}$

I am not sure what is meant by the sentence 'Experiments determined the relative weight of the volatile chloride to be about 233'. By 'volatile chloride', are they referring to zirconium oxide or something else? It seems that they are referring to something else since a chloride is a compound of chlorine with another element or group.

Also, by the statement 'assuming that zirconium in the oxide has the same valency' are they saying that zirconium has the same valency as chlorine (i.e. -1)?

Any insights are appreciated.

Answer 1

First, let's solve the problem. Both relative atomic weight $A_\mathrm{r}$ and relative molecular weight $M_\mathrm{r}$ are a historical dimensionless terms denoting relative atomic mass and molecular mass, respectively:

$$A_\mathrm{r} = \frac{m_\mathrm{a}}{m_\mathrm{u}}$$

$$M_\mathrm{r} = \frac{M_\mathrm{B}}{M_\mathrm{u}}$$

where $m$ corresponds to atomic mass, $M$ – to molecular mass, and index $\mathrm{u}$ refers to unified atomic mass unit ($= m_\mathrm{a}(\ce{^{12}C})/12$). Since $\ce{Cl}$ in chlorides is monovalent, then formula of unknown zirconium chloride is $\ce{ZrCl_x}$, where $x$ coincides with the valency of $\ce{Zr}$. From this the relative weight of $\ce{ZrCl_x}$ is

$$M_\mathrm{r}(\ce{\ce{ZrCl_x}}) = A_\mathrm{r}(\ce{Zr}) + xA_\mathrm{r}(\ce{Cl})$$

relative atomic masses of both elements were known ($A_\mathrm{r}(\ce{Zr})\approx 91$, $A_\mathrm{r}(\ce{Cl})\approx 36$), so that the valency of zirconium can be found via $x$:

$$91 + 36x = 233 \implies x \approx 4$$

Since oxygen in oxides has valency of $2$, then the formula of unknown zirconium oxide is $\ce{ZrO2}$.

By 'volatile chloride', are they referring to zirconium oxide or something else?

No, they refer to zirconium chloride $\ce{ZrCl4}$. They mention volatility probably to underline how zirconium and hafnium are separated — by differences in sublimation of their volatile compounds.

Also, by the statement 'assuming that zirconium in the oxide has the same valency' are they saying that zirconium has the same valency as chlorine (i.e. -1)?

No, this means that zirconium in both oxide $\ce{ZrO2}$ and chloride $\ce{ZrCl4}$ has the same valency.

Answer 2

It is always helpful if you also cite the reference from which you are quoting the question. The "relative weight" is old terminology which you will find in 19th-century chemistry textbooks which is the same as molecular weight. Since nobody could determine absolute weights (and the practice still continues), the term relative is added. All atomic weights quoted today are also relative but the word relative is often omitted today and perhaps taken for granted. I am adding a screenshot of an early 20th-century book, Introduction to inorganic chemistry By Alexander Smith

The second part "Assuming that zirconium in the oxide has the same valency" simply implies that the oxidation state of Zr in the chloride is the same as the oxide. This helped the chemists in the laws of definite proportions.