Problem 87
Question
Calculate the molar mass of each of these compounds and the mass percent of each element. (a) PbS, lead(II) sulfide, galena (b) \(\mathrm{C}_{2} \mathrm{H}_{6},\) ethane, a hydrocarbon fuel (c) \(\mathrm{CH}_{3} \mathrm{COOH}\), acetic acid, an important ingredient in vinegar (d) \(\mathrm{NH}_{4} \mathrm{NO}_{3}\), ammonium nitrate, a fertilizer
Step-by-Step Solution
Verified Answer
PbS: 239.27 g/mol; Pb 86.62%, S 13.38%. C₂H₆: 30.07 g/mol; C 79.88%, H 20.12%. CH₃COOH: 60.05 g/mol; C 40.01%, H 6.71%, O 53.28%. NH₄NO₃: 80.052 g/mol; N 35.00%, H 5.04%, O 59.96%.
1Step 1: Calculate Molar Mass of PbS
To find the molar mass of PbS (lead(II) sulfide), add the atomic masses of lead (Pb) and sulfur (S). The atomic mass of Pb is approximately 207.2 g/mol and for S, it is approximately 32.07 g/mol.\[\text{Molar Mass of PbS} = 207.2\, ext{g/mol} + 32.07\, ext{g/mol} = 239.27\, ext{g/mol}\]
2Step 2: Calculate Mass Percent for PbS
The mass percent of each element in PbS can be calculated using the formula: \[\text{Mass Percent of Element} = \left(\frac{\text{Mass of Element in 1 mol of compound}}{\text{Molar Mass of compound}}\right) \times 100\%\]For lead (Pb):\[\text{Mass Percent of Pb} = \left(\frac{207.2}{239.27}\right) \times 100\% \approx 86.62\%\]For sulfur (S):\[\text{Mass Percent of S} = \left(\frac{32.07}{239.27}\right) \times 100\% \approx 13.38\%\]
3Step 3: Calculate Molar Mass of \\ \mathrm{C}_{2} \mathrm{H}_{6}
For ethane, calculate the molar mass by summing the atomic masses of its constituent elements: carbon (C) and hydrogen (H). The atomic mass of C is approximately 12.01 g/mol and H is approximately 1.008 g/mol.\[\text{Molar Mass} = 2(12.01\, ext{g/mol}) + 6(1.008\, ext{g/mol}) = 24.02\, ext{g/mol} + 6.048\, ext{g/mol} = 30.07\, ext{g/mol}\]
4Step 4: Calculate Mass Percent for \\ \mathrm{C}_{2} \mathrm{H}_{6}
For ethane, the mass percent of each element is calculated using the formula:For carbon (C):\[\text{Mass Percent of C} = \left(\frac{24.02}{30.07}\right) \times 100\% \approx 79.88\%\]For hydrogen (H):\[\text{Mass Percent of H} = \left(\frac{6.048}{30.07}\right) \times 100\% \approx 20.12\%\]
5Step 5: Calculate Molar Mass of \\ \mathrm{CH}_{3} \mathrm{COOH}
For acetic acid, sum the atomic masses of all its constituent atoms: 2 carbon (C) atoms, 4 hydrogen (H) atoms, and 2 oxygen (O) atoms.\[\text{Molar Mass} = 2(12.01\, ext{g/mol}) + 4(1.008\, ext{g/mol}) + 2(16.00\, ext{g/mol})\]\[= 24.02\, ext{g/mol} + 4.032\, ext{g/mol} + 32.00\, ext{g/mol} = 60.05\, ext{g/mol}\]
6Step 6: Calculate Mass Percent for \\ \mathrm{CH}_{3} \mathrm{COOH}
To find the mass percent of each element:For carbon (C):\[\text{Mass Percent of C} = \left(\frac{24.02}{60.05}\right) \times 100\% \approx 40.01\%\]For hydrogen (H):\[\text{Mass Percent of H} = \left(\frac{4.032}{60.05}\right) \times 100\% \approx 6.71\%\]For oxygen (O):\[\text{Mass Percent of O} = \left(\frac{32.00}{60.05}\right) \times 100\% \approx 53.28\%\]
7Step 7: Calculate Molar Mass of \\ \mathrm{NH}_{4} \mathrm{NO}_{3}
Sum the atomic masses: 2 nitrogen (N) atoms, 4 hydrogen (H) atoms, and 3 oxygen (O) atoms.\[\text{Molar Mass} = 2(14.01\, ext{g/mol}) + 4(1.008\, ext{g/mol}) + 3(16.00\, ext{g/mol})\]\[= 28.02\, ext{g/mol} + 4.032\, ext{g/mol} + 48.00\, ext{g/mol} = 80.052\, ext{g/mol}\]
8Step 8: Calculate Mass Percent for \\ \mathrm{NH}_{4} \mathrm{NO}_{3}
Find the mass percent of each element: For nitrogen (N):\[\text{Mass Percent of N} = \left(\frac{28.02}{80.052}\right) \times 100\% \approx 35.00\%\]For hydrogen (H):\[\text{Mass Percent of H} = \left(\frac{4.032}{80.052}\right) \times 100\% \approx 5.04\%\]For oxygen (O):\[\text{Mass Percent of O} = \left(\frac{48.00}{80.052}\right) \times 100\% \approx 59.96\%\]
Key Concepts
Mass Percent CalculationChemical Formula InterpretationAtomic Mass Usage
Mass Percent Calculation
Mass percent is a useful way to express how much of each element is present in a compound. It is given as a percentage of the total mass of the compound. To calculate mass percent, you need to know both the mass of each element in one mole of the compound and the molar mass of the entire compound.
Here's how you can do it:
Here's how you can do it:
- First, find the molar mass of the compound by adding the atomic masses of all elements in the compound.
- Then, calculate the mass of each element in one mole of the compound. This is simply the atomic mass of the element multiplied by the number of atoms of that element in the formula.
- Next, use the formula for mass percent: \( \text{Mass Percent of Element} = \left(\frac{\text{Mass of Element in 1 mol of compound}}{\text{Molar Mass of compound}}\right) \times 100\% \).
Chemical Formula Interpretation
Understanding chemical formulas is like knowing a secret language that tells you about the composition of substances. Each chemical formula reveals which elements are present and how many atoms of each are contained in a single molecule of the compound.
Let's decode how to interpret these formulas:
Let's decode how to interpret these formulas:
- Each symbol in a chemical formula represents a different element from the periodic table.
- The subscript number following each element indicates how many atoms of that element are in one molecule. For instance, \(\mathrm{CH}_{3}\mathrm{COOH}\) has two carbons, four hydrogens, and two oxygens.
- If there is no subscript, it means that there is only one atom of that element.
Atomic Mass Usage
Atomic mass plays a crucial role in many chemistry calculations, including determining the molar mass of compounds. Each element's atomic mass is based on the unified atomic mass unit and reflects the average mass of atoms, considering isotopes of the element.
Here's how you can effectively use atomic mass:
Here's how you can effectively use atomic mass:
- Find the atomic mass of each element on the periodic table, usually expressed in grams per mole (g/mol).
- Multiply the atomic mass of each element by the number of atoms present in the compound to find the total mass for that element.
- Add these values together to get the molar mass of the compound.
Other exercises in this chapter
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