Question

A 0.750 g sample of tin is oxidized with nitric acid to form tin oxide. Calculate the

Answer 1

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:

$mSn_xO_y = mSn + mO = 0.750 g + 0.21 g = 0.96 g$

Then, we will calculate the mass percent of each element in the oxide.

$\% Sn = \frac{0.750g}{0.96g} \times 100\% = 78\%\\\\\% O = \frac{0.21g}{0.96g} \times 100\% = 22\%$

Next, we will divide each percentage by the atomic mass of the element.

$Sn: 78/118.71 = 0.66\\\\O: 22/16.00 = 1.4$

Finally, we will divide both numbers by the smallest one, i.e. 0.66.

$Sn: 0.66/0.66 = 1\\\\O: 1.4/0.66 \approx 2$

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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