Chapter 10: Problem 55
Which has the larger percent by mass of sulfur, \(\mathrm{H}_{2} \mathrm{SO}_{3}\) or \(\mathrm{H}_{2} \mathrm{S}_{2} \mathrm{O}_{8} ?\)
Short Answer
Expert verified
The compound \(\mathrm{H}_{2}\mathrm{SO}_{3}\) has a larger percent by mass of sulfur (39.04%) than \(\mathrm{H}_{2} \mathrm{S}_{2} \mathrm{O}_{8}\) (36.03%).
Step by step solution
01
Find molar mass of each compound
First, find the molar mass of the two compounds: \(\mathrm{H}_{2}\mathrm{SO}_{3}\) and \(\mathrm{H}_{2} \mathrm{S}_{2} \mathrm{O}_{8}\).
For \(\mathrm{H}_{2}\mathrm{SO}_{3}\), the molar mass of H is 1.01 g/mol, S is 32.07 g/mol, and O is 16.00 g/mol. So, the molar mass is:
\(2 \times 1.01 + 1 \times 32.07 + 3 \times 16.00 =\) 82.08 g/mol.
For \(\mathrm{H}_{2} \mathrm{S}_{2} \mathrm{O}_{8}\), the molar mass is:
\(2 \times 1.01 + 2 \times 32.07 + 8 \times 16.00 =\) 178.16 g/mol.
02
Calculate the mass of sulfur in each compound
Next, find the mass of sulfur in each compound.
For \(\mathrm{H}_{2}\mathrm{SO}_{3}\), there is 1 S atom per molecule, so the mass of sulfur is:
1 x 32.07 = 32.07 g/mol.
For \(\mathrm{H}_{2} \mathrm{S}_{2} \mathrm{O}_{8}\), there are 2 S atoms per molecule, so the mass of sulfur is:
2 x 32.07 = 64.14 g/mol.
03
Calculate the percent by mass of sulfur in each compound
Now, we will calculate the percent by mass of sulfur in each compound.
For \(\mathrm{H}_{2}\mathrm{SO}_{3}\), the percent by mass of sulfur is:
\(\frac{32.07}{82.08} \times 100 =\) 39.04%.
For \(\mathrm{H}_{2} \mathrm{S}_{2} \mathrm{O}_{8}\), the percent by mass of sulfur is:
\(\frac{64.14}{178.16} \times 100 =\) 36.03%.
04
Compare the sulfur mass percentages
Comparing the two percentages calculated in Step 3:
- \(\mathrm{H}_{2}\mathrm{SO}_{3}\): 39.04%;
- \(\mathrm{H}_{2} \mathrm{S}_{2} \mathrm{O}_{8}\): 36.03%.
\(\mathrm{H}_{2}\mathrm{SO}_{3}\) has a larger percent by mass of sulfur than \(\mathrm{H}_{2} \mathrm{S}_{2} \mathrm{O}_{8}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Molar Mass Calculation
Understanding molar mass is fundamental when analyzing chemical compounds. The molar mass of a molecule is the sum of the atomic masses of all the atoms present in the molecule. It is usually expressed in grams per mole (g/mol). To calculate the molar mass of a compound:
- List the elements present in the compound and their respective numbers.
- Determine the atomic mass of each element from the periodic table.
- Multiply each atomic mass by the number of atoms of the element in the compound.
- Add these values together to get the total molar mass.
Elemental Analysis
Elemental analysis involves determining the mass and the percent contribution of individual elements in a chemical compound. This is particularly useful when assessing the properties of a compound. To analyze the sulfur content in the chemical compounds \(\mathrm{H}_{2}\mathrm{SO}_{3}\) and \(\mathrm{H}_{2}\mathrm{S}_{2}\mathrm{O}_{8}\):
- First, find the number of sulfur atoms present in each compound. \(\mathrm{H}_{2}\mathrm{SO}_{3}\) contains 1 sulfur atom while \(\mathrm{H}_{2}\mathrm{S}_{2}\mathrm{O}_{8}\) contains 2 sulfur atoms.
- Multiply the atomic mass of sulfur, which is 32.07 g/mol, by the number of sulfur atoms.
- The sum gives you the total mass of sulfur present in each compound. Thus, the total mass of sulfur in \(\mathrm{H}_{2}\mathrm{SO}_{3}\) is 32.07 g/mol, whereas it is 64.14 g/mol in \(\mathrm{H}_{2}\mathrm{S}_{2}\mathrm{O}_{8}\).
Chemical Compounds Comparison
Comparing chemical compounds can offer insights into their chemical behavior and relevance. One way to compare compounds is by calculating and comparing their percent composition, which represents the ratio of the mass of each element to the total mass of the compound. For example:
- In \(\mathrm{H}_{2}\mathrm{SO}_{3}\), the percent composition of sulfur is calculated as \(\frac{32.07}{82.08} \times 100\approx 39.04\%\).
- In \(\mathrm{H}_{2}\mathrm{S}_{2}\mathrm{O}_{8}\), the percent composition of sulfur is \(\frac{64.14}{178.16} \times 100\approx 36.03\%\).