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Chapter 12: Problem 22

A solution is made by dissolving 26.42 g of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) in enough \(\mathrm{H}_{2} \mathrm{O}\) to make 50.00 \(\mathrm{mL}\) of solution. a. What is the molar mass of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) ? b. What is the molarity of this solution?

Short Answer

Expert verified
Molar mass of (NH_4)_2SO_4 = 132.17 g/mol. Molarity = 4.000 M.

Step by step solution

01

Calculate the Molar Mass of \(\text{(NH}_4\text{)}_2\text{SO}_4\)

First, determine the molar mass of \(\text{(NH}_4\text{)}_2\text{SO}_4\) by adding the atomic masses of all the atoms in the formula.Elements present in \(\text{(NH}_4\text{)}_2\text{SO}_4\):- N (Nitrogen) atoms: 2- H (Hydrogen) atoms: 8- S (Sulfur) atoms: 1- O (Oxygen) atoms: 4Atomic masses:- Nitrogen: 14.01 g/mol- Hydrogen: 1.01 g/mol- Sulfur: 32.07 g/mol- Oxygen: 16.00 g/molCalculate the molar mass:\(\text{Molar mass} = [2 \cdot 14.01] + [8 \cdot 1.01] + [1 \cdot 32.07] + [4 \cdot 16.00] = 28.02 + 8.08 + 32.07 + 64.00 = 132.17\) g/mol
02

Convert Mass to Moles

Next, use the molar mass to convert grams to moles. Given the mass of \(\text{(NH}_4\text{)}_2\text{SO}_4\) is 26.42 g.Moles of \(\text{(NH}_4\text{)}_2\text{SO}_4\) = \( \frac{26.42 \text{g}}{132.17 \text{g/mol}} \) \(\approx 0.1999 \text{mol}\)
03

Convert Volume to Liters

The volume of the solution is given as 50.00 mL. Convert this volume to liters since molarity is expressed in moles per liter.Volume in liters = \( \frac{50.00 \text{mL}}{1000 \text{mL/L}} \) \( = 0.05000 \text{L} \)
04

Calculate the Molarity

Finally, calculate the molarity of the solution using the formula:Molarity (M) = \( \frac{\text{moles of solute}}{\text{liters of solution}} \)Molarity = \( \frac{0.1999 \text{mol}}{0.05000 \text{L}} \) \( = 3.998 \text{M} \)However, considering significant figures, the final answer should be rounded to 4.000 M.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

molar mass calculation
Calculating the molar mass of a compound involves adding the atomic masses of each element present in the compound. For \(\text{(NH}_4\text{)}_2\text{SO}_4\), we need to account for: 2 nitrogen (N) atoms, 8 hydrogen (H) atoms, 1 sulfur (S) atom, and 4 oxygen (O) atoms. Here's how you can determine it:
  • First, find the atomic mass of each element. Nitrogen is 14.01 g/mol, hydrogen is 1.01 g/mol, sulfur is 32.07 g/mol, and oxygen is 16.00 g/mol.
  • Next, multiply these atomic masses by the number of atoms in the compound: \[ (2 \times 14.01) + (8 \times 1.01) + (1 \times 32.07) + (4 \times 16.00) = 28.02 + 8.08 + 32.07 + 64.00 \]
  • Finally, add these values to get the molar mass: 132.17 g/mol.
This sum is the molar mass of \(\text{(NH}_4\text{)}_2\text{SO}_4\), which is crucial for further calculations.
molarity
Molarity is a measurement of concentration, representing the number of moles of solute per liter of solution. The formula for calculating molarity (M) is: \text{Molarity} = \frac{\text{moles of solute}}{\text{liters of solution}}\. To find the molarity of our \(\text{(NH}_4\text{)}_2\text{SO}_4\) solution:
  • First, calculate the moles of solute using the given mass and molar mass. If you have 26.42 g of \(\text{(NH}_4\text{)}_2\text{SO}_4\) and a molar mass of 132.17 g/mol, then \[\text{Moles of } \text{(NH}_4\text{)}_2\text{SO}_4 = \frac{26.42 \text{ g}}{132.17 \text{ g/mol}} \approx 0.1999 \text{ mol}\].
  • Secondly, convert the volume of the solution from milliliters to liters (i.e., 50.00 mL to 0.05000 L).
  • Finally, use these values in the molarity formula: \text{Molarity} = \frac{0.1999 \text{ mol}}{0.05000 \text{ L}} = 3.998 \text{ M}\, which rounds to 4.000 M when considering significant figures.
conversion of units
Converting units is essential when performing calculations in chemistry since different measurements may be given in various units. Here are some common conversions:
  • For volume, converting milliliters (mL) to liters (L) is often necessary, since molarity is measured in moles per liter. For example, converting 50.00 mL to liters involves dividing by 1000: \[ \frac{50.00 \text{ mL}}{1000 \text{ mL/L}} = 0.05000 \text{ L} \]
  • Mass may need to be converted between grams and kilograms, where 1 kg = 1000 g. However, for most chemical solutions, we use grams (g).
Always remember to keep your units consistent across your calculations to avoid mistakes. Make sure to convert to the appropriate units before plugging numbers into formulas.
significant figures
Significant figures indicate the precision of a measurement. When performing calculations, the number of significant figures should match the least precise measurement. Here are some rules for determining significant figures:
  • All non-zero digits are significant.
  • Any zeros between significant digits are also significant.
  • Leading zeros (zeros before the first non-zero digit) are not significant.
  • Trailing zeros in a decimal number are significant.
  • In our calculation, the mass 26.42 g has 4 significant figures and the volume 50.00 mL has 4 significant figures.
When rounding your final result, ensure it matches the number of significant figures in your least precise measurement. For example, the calculated molarity of 3.998 M should be rounded to 4.000 M to reflect the original data's precision.

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