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Molar Mass of SnO2 (Tin(IV) Oxide)

What is the molar mass of SnO2 (tin(IV) oxide)?

Subject: General Chemistry Chapter: Chemical Compounds Topic: Molecular Mass and Formula Mass Answer included
molar mass of sn02 molar mass of SnO2 tin(IV) oxide molar mass formula mass SnO2 molecular mass SnO2 atomic mass tin atomic mass oxygen SnO2 molar mass calculation
Accepted answer Answer included

The phrase “molar mass of sn02” is interpreted as the molar mass of tin(IV) oxide, SnO2, where the symbol for oxygen is the letter O (not the numeral 0). The molar mass is the sum of the atomic masses in the formula, expressed in g/mol.

Chemical formula and meaning of subscripts

SnO2 contains one tin atom (Sn) and two oxygen atoms (O). The subscript 2 multiplies oxygen’s atomic mass by 2. Tin(IV) oxide is consistent with oxidation states Sn4+ and O2−, giving a neutral compound.

Atomic masses used

Element Symbol Count in SnO2 Atomic mass (g/mol) Contribution (g/mol)
Tin Sn 1 118.71 118.71
Oxygen O 2 15.999 31.998

Molar mass calculation

\[ M(\mathrm{SnO_2}) = M(\mathrm{Sn}) + 2 \times M(\mathrm{O}) \] \[ M(\mathrm{SnO_2}) = 118.71 + 2 \times 15.999 = 118.71 + 31.998 = 150.708\ \mathrm{g/mol} \]

Rounded to typical periodic-table precision, the molar mass of SnO2 is 150.71 g/mol. The same numeric value is the formula mass in atomic mass units (amu) because \(1\ \mathrm{amu}\) per particle corresponds to \(1\ \mathrm{g/mol}\) per mole.

Mass contribution picture

SnO2 composition and molar-mass contributions A schematic SnO2 molecule with one tin atom and two oxygen atoms, plus a bar chart showing the mass contributions: tin contributes 118.71 g/mol (~78.8%), oxygen contributes 31.998 g/mol (~21.2%), total 150.708 g/mol. SnO2 composition O 15.999 O 15.999 Sn 118.71 Atomic masses shown in g/mol (periodic-table averages). Mass contributions to molar mass 0 50 100 150 Sn O2 118.71 g/mol ≈ 78.8% 31.998 g/mol ≈ 21.2% Total: 150.708 g/mol
The bars show how the subscript in SnO2 doubles oxygen’s contribution. Tin contributes most of the molar mass because its atomic mass is much larger than oxygen’s.

Common rounding conventions

Periodic tables often list oxygen as 16.00 and tin as 118.71. Substituting \(16.00\) for \(15.999\) gives \(118.71 + 2 \times 16.00 = 150.71\ \mathrm{g/mol}\), matching the rounded value.

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