# Atomic formula obtained by mixing given percentages of oxides

Calculate the atomic formula of a compound which is made by $$93\,\%$$ of $$\ce{SnO2}$$ and $$7\,\%$$ of $$\ce{Sb2O5}$$.

The molar mass of $$\ce{SnO2}$$ is $$\pu{150.71 g mol^-1}$$ and the molar mass of $$\ce{Sb2O5}$$ is $$\pu{323.517 g mol^-1}$$. I can find the corresponding amounts of each substance as follows:

\begin{align} n(\ce{SnO2}) &= \frac{93\,\%}{\pu{150.71 g mol^-1}} = \pu{0.61 mol} \\ n(\ce{Sb2O5}) &= \frac{7\,\%}{\pu{323.517 g mol^-1}} = \pu{0.022 mol} \end{align}

and multiply these by the number of atoms that I have obtaining $$0.61~\ce{Sn}$$, $$1.24~\ce{O}$$ for $$\ce{SnO2}$$, and $$0.043~\ce{Sb}$$, $$0.108~\ce{O}$$ for $$\ce{Sb2O5}$$.

Finally, summing similar atoms coefficients for oxygen, what I get is

$$\ce{Sn_{0.62}Sb_{0.043}O_{1.35}},$$

and choosing $$x=0.5$$ and $$y=2,$$ the formula becomes $$\ce{Sn_{0.5}Sb2O6}.$$ Is this correct?

• Better is to compute molar amount ratios for elements. $\ce{Sn_xSb_yO_{2x+(5/2)y}}$ Feb 21, 2022 at 14:09
• thanks, could you elaborate on the example please? Feb 21, 2022 at 14:11
• Useful links for text and formula formatting (not to be applied to titles): Notation basics , Formatting of math/chem expressions and upright vs italic Feb 21, 2022 at 14:19
• I know arithmetics, I just didn't understand what you meant, to calculate it more generally or what. So I just asked to be more explicit on your comment Feb 21, 2022 at 14:20
• We say (molar) amount of X, not moles (of) X, like we say mass of X, not kilograms of X. Note that the mole is the unit like kilogram, mol is unit symbol like kg. It is often confused. // Yes, it is similar like normalizing of chemical equation coefficients to have the smallest possible integers ( what is not possible here.) Feb 21, 2022 at 14:34

Technically, your approach is in general correct. There is, however, no indication that the compound should be stoichiometric, and there are additional less obvious details you can use to prove this.

First, you are not presented with the data elemental analysis, rather composition expressed in terms of oxides of the elements in their highest oxidation states, which usually means either of the two scenarios:

1. it's data obtained with DTA or another thermoanalytical method;
2. it's the precursors used for high-temperature synthesis (homogeneous, high-purity oxides usually contain fully oxidized elements).

What should be taken from this is that the amount of oxygen cannot be determined from its content in given oxides, and that the elements in the target compound could be reduced.

Second, pay attention to the numbers: $$7\,\%$$ of the relatively heavy element's oxide $$\ce{Sb2O5}$$ is likely a dopant for $$\ce{SnO2}.$$

Third, I checked ICSD database for $$\ce{Sn-Sb-O}$$ system and all entries (24) deposited prior to 2017 (I don't have newer version) are for the defect rutile structure of tin-antimony oxide $$\ce{Sn_xSb_{1-x}O2},$$ a phase of variable composition which sustains cassiterite antimony-doped structure [1, 2].

Having found the formula unit $$\ce{Sn_xSb_{1-x}O2}$$ by inference and literature research, it is now possible to compose an equation with one unknown:

\begin{align} \frac{x}{1 - x} &= \frac{w(\ce{SnO2})}{M(\ce{SnO2)}}\cdot\frac{M(\ce{Sb2O5})}{2\cdot w(\ce{Sb2O5})} \\ &= \frac{93\,\%}{\pu{150.71 g mol^-1}}\times\frac{\pu{323.52 g mol^-1}}{2\times 7\,\%} \\ &= 14.26 \\ \implies x &\approx 0.93 \end{align}

and propose the formula $$\ce{Sn_{0.93}Sb_{0.07}O2}.$$ The coincidence of the composition with the coefficients of the formula unit is accidental and is due to the numerical values of the molar masses.

### References

1. Berry, F. J.; Greaves, C. A Neutron Diffraction Investigation of the Defect Rutile Structure of Tin–Antimony Oxide. J. Chem. Soc., Dalton Trans. 1981, 12, 2447–2451. DOI: 10.1039/DT9810002447.
2. Tena, M. A.; Sorlí, S.; Llusar, M.; Badenes, J. A.; Forés, A.; Monrós, G. Study of Sb-doped SnO2 Gray Ceramic Pigment with Cassiterite Structure. Z. anorg. allg. Chem. 2005, 631 (11), 2188–2191. DOI: 10.1002/zaac.200570038.

The only reliable formula is probably made with $$\ce{28 SnO2 + 1 Sb2O5}$$, if it exists.

$$28$$ mol $$\ce{SnO2}$$ weighs $$28·150.7 = 4220$$ g. One mole $$\ce{Sb2O5}$$ weighs $$323.5$$ g. So the total weighs $$4543.5$$ g. The proportion of $$\ce{SnO2}$$ is $$\frac{4220}{4543.5} = 0.929 = 93$$%.