A 0.750 g sample of tin is oxidized with nitric acid to form tin oxide. if the original tin sample gained 0.21 g of oxygen, the empirical formula of tin oxide is SnO₂.
A 0.750 g sample of tin is oxidized with nitric acid to form tin oxide. if the original tin sample gained 0.21 g of oxygen, the mass of tin oxide is:
[tex]mSn_xO_y = mSn + mO = 0.750 g + 0.21 g = 0.96 g[/tex]
Then, we will calculate the mass percent of each element in the oxide.
[tex]\% Sn = \frac{0.750g}{0.96g} \times 100\% = 78\%\\\\\% O = \frac{0.21g}{0.96g} \times 100\% = 22\%[/tex]
Next, we will divide each percentage by the atomic mass of the element.
[tex]Sn: 78/118.71 = 0.66\\\\O: 22/16.00 = 1.4[/tex]
Finally, we will divide both numbers by the smallest one, i.e. 0.66.
[tex]Sn: 0.66/0.66 = 1\\\\O: 1.4/0.66 \approx 2[/tex]
The empirical formula of tin oxide is SnO₂.
A 0.750 g sample of tin is oxidized with nitric acid to form tin oxide. if the original tin sample gained 0.21 g of oxygen, the empirical formula of tin oxide is SnO₂.
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