Step-by-Step Guidance for CHEM 111 Exam #5 Questions

Q1. How many atoms of carbon are there in 0.37 mol of procaine, C13H20N2O2?

Background

Topic: Mole-to-Atom Conversions

This question tests your ability to convert between moles of a compound and the number of atoms of a specific element within that compound using Avogadro's number.

Key Terms and Formulas

• Mole: A unit representing $6.022 \times 10^{23}$ entities (Avogadro's number).

• Avogadro's Number: $6.022 \times 10^{23}$ particles/mol.

• Procaine formula: C13H20N2O2 (13 carbon atoms per molecule).

Step-by-Step Guidance

1. Start by identifying how many carbon atoms are in one molecule of procaine (from the formula, there are 13).

2. Calculate the number of molecules in 0.37 mol using Avogadro's number: $\text{Number of molecules} = 0.37 \text{ mol} \times 6.022 \times 10^{23} \text{ molecules/mol}$

3. Multiply the number of molecules by the number of carbon atoms per molecule to get the total number of carbon atoms: $\text{Total C atoms} = \text{Number of molecules} \times 13$

Try solving on your own before revealing the answer!

Final Answer: 2.9 × 1024 atoms of C

We multiplied the number of molecules by 13 to get the total number of carbon atoms in 0.37 mol of procaine.

Q2. How many molecules are there in 3.24 g of nicotine, C10H14N2?

Background

Topic: Mass-to-Molecule Conversions

This question tests your ability to convert a mass of a compound to the number of molecules using molar mass and Avogadro's number.

Key Terms and Formulas

• Molar Mass: The mass of one mole of a substance (g/mol).

• Avogadro's Number: $6.022 \times 10^{23}$ molecules/mol.

• Nicotine formula: C10H14N2

Step-by-Step Guidance

1. Calculate the molar mass of nicotine by adding the atomic masses of all atoms in the formula.

2. Convert 3.24 g of nicotine to moles: $\text{Moles} = \frac{3.24 \text{ g}}{\text{Molar mass of nicotine}}$

3. Convert moles to molecules using Avogadro's number: $\text{Molecules} = \text{Moles} \times 6.022 \times 10^{23}$

Try solving on your own before revealing the answer!

Final Answer: 1.20 × 1022 molecules

After converting grams to moles and multiplying by Avogadro's number, you get the total number of molecules.

Q3. Using "Like Dissolves Like", determine whether carbon tetrachloride is soluble in ethanol (a polar solvent).

Background

Topic: Solubility and Intermolecular Forces

This question tests your understanding of the "like dissolves like" principle, which states that polar solvents dissolve polar solutes, and nonpolar solvents dissolve nonpolar solutes.

Key Terms

• Polar: Molecules with uneven charge distribution.

• Nonpolar: Molecules with even charge distribution.

• Carbon tetrachloride (CCl4): Nonpolar.

• Ethanol: Polar solvent.

Step-by-Step Guidance

1. Identify the polarity of carbon tetrachloride (CCl4).

2. Recall the polarity of ethanol.

3. Apply the "like dissolves like" rule to predict solubility.

Try solving on your own before revealing the answer!

Final Answer: No, carbon tetrachloride is not soluble in ethanol.

Nonpolar solutes do not dissolve well in polar solvents.

Q4. How many moles of oxygen atoms are in 21.4 mol of TNT, C7H5N3O6?

Background

Topic: Mole-to-Atom Conversions in Compounds

This question tests your ability to determine the number of moles of a specific atom in a given number of moles of a compound.

Key Terms and Formulas

• TNT formula: C7H5N3O6 (6 oxygen atoms per molecule).

• 1 mol of compound contains Avogadro's number of molecules.

Step-by-Step Guidance

1. Identify the number of oxygen atoms in one molecule of TNT (6 per molecule).

2. Multiply the number of moles of TNT by the number of oxygen atoms per molecule: $\text{Moles of O atoms} = 21.4 \text{ mol TNT} \times 6$

Try solving on your own before revealing the answer!

Final Answer: 128 moles of O atoms

Each mole of TNT contains 6 moles of oxygen atoms, so multiply accordingly.

Q5. How many moles of nitrogen atoms are in 21.4 mol of TNT, C7H5N3O6?

Background

Topic: Mole-to-Atom Conversions in Compounds

This question is similar to the previous one but focuses on nitrogen atoms.

Key Terms and Formulas

• TNT formula: C7H5N3O6 (3 nitrogen atoms per molecule).

Step-by-Step Guidance

1. Identify the number of nitrogen atoms in one molecule of TNT (3 per molecule).

2. Multiply the number of moles of TNT by the number of nitrogen atoms per molecule: $\text{Moles of N atoms} = 21.4 \text{ mol TNT} \times 3$

Try solving on your own before revealing the answer!

Final Answer: 64.2 moles of N atoms

Each mole of TNT contains 3 moles of nitrogen atoms, so multiply accordingly.

Q6. What mass of chlorine gas is required to react completely with 72.5 g of aluminum? (2Al(s) + 3Cl2(g) → 2AlCl3(s))

Background

Topic: Stoichiometry (Mass-to-Mass Conversions)

This question tests your ability to use stoichiometry to relate the mass of one reactant to the mass of another in a balanced chemical equation.

Key Terms and Formulas

• Molar mass of Al: 26.98 g/mol

• Molar mass of Cl2: 70.90 g/mol

• Stoichiometric coefficients: 2 mol Al : 3 mol Cl2

Step-by-Step Guidance

1. Convert 72.5 g of Al to moles: $\text{Moles Al} = \frac{72.5 \text{ g}}{26.98 \text{ g/mol}}$

2. Use the stoichiometric ratio from the balanced equation to find moles of Cl2 needed: $\text{Moles Cl}_2 = \text{Moles Al} \times \frac{3 \text{ mol Cl}_2}{2 \text{ mol Al}}$

3. Convert moles of Cl2 to grams: $\text{Mass Cl}_2 = \text{Moles Cl}_2 \times 70.90 \text{ g/mol}$

Try solving on your own before revealing the answer!

Final Answer: 286 g of Cl2

By converting grams of Al to moles, using the stoichiometric ratio, and then converting to grams of Cl2, you find the required mass.

Q7. Calculate the mass, in grams, of 8.35 × 1022 molecules of CBr4.

Background

Topic: Molecule-to-Mass Conversions

This question tests your ability to convert a number of molecules to mass using Avogadro's number and molar mass.

Key Terms and Formulas

• Avogadro's Number: $6.022 \times 10^{23}$ molecules/mol

• Molar mass of CBr4: (12.01 + 4 × 79.90) g/mol

Step-by-Step Guidance

1. Calculate the molar mass of CBr4.

2. Convert molecules to moles: $\text{Moles} = \frac{8.35 \times 10^{22}}{6.022 \times 10^{23}}$

3. Convert moles to grams: $\text{Mass} = \text{Moles} \times \text{Molar mass of CBr}_4$

Try solving on your own before revealing the answer!

Final Answer: 46.0 g

After converting molecules to moles and multiplying by the molar mass, you get the mass in grams.

Q8. The solubility of KBr is 80g/100g of water. Calculate the following solution solubility: 10.83 g KBr is added to 38.84 g of water.

Background

Topic: Solution Concentration and Solubility

This question tests your ability to compare the amount of solute added to the solvent with the solubility limit, and to calculate the concentration in g/100g water.

Key Terms and Formulas

• Solubility: Maximum amount of solute that can dissolve in a given amount of solvent at a specific temperature.

• Concentration (g/100g water): $\frac{\text{grams solute}}{\text{grams water}} \times 100$

Step-by-Step Guidance

1. Calculate the concentration of KBr in the solution: $\text{Concentration} = \frac{10.83 \text{ g}}{38.84 \text{ g}} \times 100$

2. Compare this value to the solubility limit (80g/100g water) to determine if all the KBr will dissolve.

Try solving on your own before revealing the answer!

Final Answer: 27.88 g/100g water

This value is below the solubility limit, so all the KBr will dissolve.

Q9. Consider the fermentation reaction of glucose: C6H12O6 → 2C2H5OH + 2CO2. If 180.2 g of C6H12O6 was placed in a vat with yeast and 67.7 g of C2H5OH was obtained, what was the percent yield of C2H5OH?

Background

Topic: Percent Yield in Chemical Reactions

This question tests your ability to calculate percent yield from actual and theoretical yields.

Key Terms and Formulas

• Theoretical yield: Maximum amount of product possible from given reactants.

• Percent yield: $\frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100$

• Molar mass of glucose (C6H12O6): 180.2 g/mol

• Molar mass of ethanol (C2H5OH): 46.07 g/mol

Step-by-Step Guidance

1. Convert 180.2 g of glucose to moles: $\text{Moles glucose} = \frac{180.2 \text{ g}}{180.2 \text{ g/mol}}$

2. Use the stoichiometry of the reaction to find moles of ethanol produced per mole of glucose.

3. Calculate the theoretical yield of ethanol in grams.

4. Calculate percent yield: $\text{Percent yield} = \frac{67.7 \text{ g}}{\text{Theoretical yield}} \times 100$

Try solving on your own before revealing the answer!

Final Answer: 73.5%

The percent yield is calculated by dividing the actual yield by the theoretical yield and multiplying by 100.

Q10. Aluminum oxide, Al2O3, is used as a filler for paints and varnishes. Calculate the number of moles in 47.51 g of Al2O3.

Background

Topic: Mass-to-Mole Conversions

This question tests your ability to convert mass to moles using molar mass.

Key Terms and Formulas

• Molar mass of Al2O3: (2 × 26.98) + (3 × 16.00) g/mol

• Moles: $\text{Moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}$

Step-by-Step Guidance

1. Calculate the molar mass of Al2O3.

2. Divide the given mass by the molar mass to find moles: $\text{Moles} = \frac{47.51}{\text{Molar mass of Al}_2\text{O}_3}$

Try solving on your own before revealing the answer!

Final Answer: 0.4660 mol

By dividing the mass by the molar mass, you get the number of moles.

Q11. If 3.00 moles of NH3 and 4.00 moles of O2 react in the following reaction, how many moles of H2O will be formed? 4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (g)

Background

Topic: Limiting Reactant and Stoichiometry

This question tests your ability to determine the limiting reactant and use stoichiometry to calculate the amount of product formed.

Key Terms and Formulas

• Limiting reactant: The reactant that is completely consumed first.

• Stoichiometric ratios from the balanced equation.

Step-by-Step Guidance

1. Calculate the mole ratio for each reactant based on the balanced equation.

2. Determine which reactant is limiting by comparing the available moles to the required stoichiometric ratio.

3. Use the limiting reactant to calculate the moles of H2O produced: $\text{Moles H}_2\text{O} = \text{Moles limiting reactant} \times \frac{6 \text{ mol H}_2\text{O}}{\text{stoichiometric coefficient of limiting reactant}}$

Try solving on your own before revealing the answer!

Final Answer: 4.50 moles

After identifying the limiting reactant and applying the stoichiometric ratio, you find the moles of water formed.

Q12. How many grams of Cl2 can be prepared from the reaction of 16.0 g of MnO2 and 30.0 g of HCl according to the following balanced chemical equation? MnO2 + 4HCl → MnCl2 + Cl2 + 2H2O

Background

Topic: Limiting Reactant and Mass-to-Mass Stoichiometry

This question tests your ability to identify the limiting reactant and calculate the mass of product formed.

Key Terms and Formulas

• Molar mass of MnO2: 86.94 g/mol

• Molar mass of HCl: 36.46 g/mol

• Molar mass of Cl2: 70.90 g/mol

• Stoichiometry from the balanced equation.

Step-by-Step Guidance

1. Convert the masses of MnO2 and HCl to moles.

2. Determine the limiting reactant by comparing the mole ratios required by the balanced equation.

3. Use the limiting reactant to calculate the moles of Cl2 produced.

4. Convert moles of Cl2 to grams: $\text{Mass Cl}_2 = \text{Moles Cl}_2 \times 70.90 \text{ g/mol}$

Try solving on your own before revealing the answer!

Final Answer: 13.0 g

After identifying the limiting reactant and converting to grams, you get the mass of Cl2 produced.

Q13. If 70.27 g of ethylene reacts with excess water to form ethanol, what is the percent yield if only 51.8 g of ethanol are actually formed? (MM of C2H4 = 28.06 g/mol; MM of H2O = 18.02 g/mol; MM of C2H5OH = 46.08 g/mol)

Background

Topic: Percent Yield in Chemical Reactions

This question tests your ability to calculate percent yield from actual and theoretical yields using molar masses and stoichiometry.

Key Terms and Formulas

• Theoretical yield: Maximum possible amount of product.

• Percent yield: $\frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100$

Step-by-Step Guidance

1. Convert 70.27 g of ethylene to moles: $\text{Moles C}_2\text{H}_4 = \frac{70.27}{28.06}$

2. Use the stoichiometry of the reaction to find moles of ethanol produced per mole of ethylene.

3. Calculate the theoretical yield of ethanol in grams.

4. Calculate percent yield: $\text{Percent yield} = \frac{51.8}{\text{Theoretical yield}} \times 100$

Try solving on your own before revealing the answer!

Final Answer: 73.8

Percent yield is calculated by dividing the actual yield by the theoretical yield and multiplying by 100.

Q14. What is the coefficient of H2O when the following equation is properly balanced? PCl3(l) + H2O(l) → H3PO3(aq) + HCl(aq)

Background

Topic: Balancing Chemical Equations

This question tests your ability to balance chemical equations and identify the correct coefficients for each compound.

Key Terms and Formulas

• Balancing equations: Ensuring the same number of each type of atom on both sides of the equation.

Step-by-Step Guidance

1. Write out the unbalanced equation and count the number of each atom on both sides.

2. Adjust coefficients to balance phosphorus, chlorine, hydrogen, and oxygen atoms.

3. Check your work to ensure all atoms are balanced.

Try solving on your own before revealing the answer!

Final Answer: 3

When balanced, the coefficient of H2O is 3.