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General Chemistry I: Mole, Molar Mass, Balancing Equations, and Limiting Reactants – Guided Study

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Q1. How many moles are in 22.5 kg Fe(NO3)2?

Background

Topic: The Mole and Molar Mass

This question tests your ability to convert mass to moles using the molar mass of a compound.

Key Terms and Formulas

  • Mole: A unit representing entities (Avogadro's number).

  • Molar Mass: The mass (in grams) of one mole of a substance.

Key formula:

Step-by-Step Guidance

  1. Convert the mass from kilograms to grams. (Remember: .)

  2. Calculate the molar mass of Fe(NO3)2 by adding the atomic masses of Fe, N, and O in the correct proportions.

  3. Set up the calculation using the formula above, plugging in the mass in grams and the calculated molar mass.

  4. Stop here and prepare to perform the division to find the number of moles.

Try solving on your own before revealing the answer!

Final Answer: 140 mol

First, convert 22.5 kg to grams: .

Molar mass of Fe(NO3)2 = .

Number of moles: moles.

So, there are approximately 125 moles of Fe(NO3)2 in 22.5 kg.

Q2. A sugar crystal contains approximately sucrose (C12H22O11) molecules. What is its mass in mg?

Background

Topic: Mole Concept and Mass-Molecule Conversions

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

Key Terms and Formulas

  • Avogadro's Number: molecules/mol

  • Molar Mass of Sucrose: Calculate using atomic masses (C, H, O)

Key formulas:

Step-by-Step Guidance

  1. Calculate the number of moles of sucrose using Avogadro's number.

  2. Determine the molar mass of sucrose (C12H22O11).

  3. Multiply the number of moles by the molar mass to get the mass in grams.

  4. Convert the mass from grams to milligrams.

Try solving on your own before revealing the answer!

Final Answer: 9.7 × 10-5 mg

Number of moles: mol.

Molar mass of sucrose: g/mol.

Mass in grams: g.

Mass in mg: mg.

Q3. A drop of water has a volume of approximately mL. How many water molecules does it contain? The density of water is 1.0 g/cm3.

Background

Topic: Mole Concept, Density, and Molecule Counting

This question tests your ability to use density to find mass, convert mass to moles, and then to number of molecules.

Key Terms and Formulas

  • Density:

  • Molar Mass of H2O: 18.02 g/mol

  • Avogadro's Number: molecules/mol

Step-by-Step Guidance

  1. Convert the volume from mL to cm3 (1 mL = 1 cm3).

  2. Use the density to calculate the mass of the water drop.

  3. Convert the mass to moles using the molar mass of water.

  4. Multiply the number of moles by Avogadro's number to find the number of molecules.

Try solving on your own before revealing the answer!

Final Answer: 2.3 × 1021 molecules

Volume: mL = $7 \times 10^{-2}$ cm3.

Mass: g/cm3 cm3 = g.

Moles: mol.

Molecules: molecules.

Q4. What is the difference between a homogeneous and a heterogeneous mixture?

Background

Topic: Classification of Matter

This question tests your understanding of the types of mixtures in chemistry.

Key Terms

  • Homogeneous mixture: A mixture with uniform composition throughout.

  • Heterogeneous mixture: A mixture with non-uniform composition; different parts can be distinguished.

Step-by-Step Guidance

  1. Define what a homogeneous mixture is and give an example (e.g., salt water).

  2. Define what a heterogeneous mixture is and give an example (e.g., salad, sand in water).

  3. Compare the two definitions, focusing on uniformity and visibility of different components.

Try explaining the difference in your own words before revealing the answer!

Final Answer:

A homogeneous mixture has the same composition throughout and looks uniform (e.g., air, salt water). A heterogeneous mixture has visibly different parts or phases (e.g., oil and water, salad). The key difference is whether the composition is uniform or not.

Q5. Classify each of the following changes as physical or chemical.

  • a. Sugar dissolves in water

  • b. A platinum ring becomes dull because of continued abrasion.

  • c. Sugar burns when heated on a skillet.

  • d. A silver surface becomes tarnished after exposure to air for a long period of time.

Background

Topic: Physical vs. Chemical Changes

This question tests your ability to distinguish between changes that alter the identity of a substance (chemical) and those that do not (physical).

Key Terms

  • Physical change: A change in state or appearance without changing chemical composition.

  • Chemical change: A change that results in the formation of new substances.

Step-by-Step Guidance

  1. For each scenario, ask: Is the substance's chemical identity changing, or just its form/appearance?

  2. Recall that dissolving, phase changes, and abrasion are usually physical, while burning and tarnishing involve chemical reactions.

  3. Classify each change accordingly, but stop before listing the answers for all parts.

Try classifying each change before revealing the answer!

Final Answer:

  • a. Physical (dissolving does not change chemical identity)

  • b. Physical (abrasion is a physical process)

  • c. Chemical (burning changes sugar to new substances)

  • d. Chemical (tarnishing involves a chemical reaction with air)

Q6. Explain the difference between Z (the atomic number) and A (the mass number).

Background

Topic: Atomic Structure

This question tests your understanding of atomic number and mass number in the context of isotopes and elements.

Key Terms

  • Atomic number (Z): Number of protons in the nucleus of an atom.

  • Mass number (A): Total number of protons and neutrons in the nucleus.

Step-by-Step Guidance

  1. Define atomic number (Z) and explain its significance for element identity.

  2. Define mass number (A) and explain how it relates to isotopes.

  3. Compare and contrast the two numbers, focusing on what each represents.

Try explaining the difference in your own words before revealing the answer!

Final Answer:

Z (atomic number) is the number of protons in an atom and determines the element. A (mass number) is the sum of protons and neutrons in the nucleus. Isotopes of an element have the same Z but different A.

Q7. Show how you can determine the molecular formula for acetic acid from the empirical formula CH2O and the added information that the molar mass of acetic acid is 60.05 g/mol.

Background

Topic: Empirical and Molecular Formulas

This question tests your ability to use empirical formula and molar mass to find the molecular formula.

Key Terms and Formulas

  • Empirical formula: Simplest whole-number ratio of atoms in a compound.

  • Molecular formula: Actual number of atoms of each element in a molecule.

Key formula:

Step-by-Step Guidance

  1. Calculate the empirical formula mass of CH2O.

  2. Divide the given molar mass by the empirical formula mass to find n (the multiple).

  3. Multiply the subscripts in the empirical formula by n to get the molecular formula.

  4. Stop here and set up the multiplication for the student to complete.

Try solving on your own before revealing the answer!

Final Answer: C2H4O2

Empirical formula mass: g/mol.

n = .

Molecular formula: .

Q8. A sample of baking soda was found to consist of 9.122 g Na, 0.4000 g H, 4.766 g C, and 19.04 g O. Calculate the moles of each element present, and determine the empirical formula of baking soda.

Background

Topic: Empirical Formula Determination

This question tests your ability to convert mass to moles for each element and determine the simplest whole-number ratio.

Key Terms and Formulas

  • Empirical formula: Simplest ratio of atoms in a compound.

  • Moles:

Step-by-Step Guidance

  1. Calculate the moles of each element using their respective atomic masses (Na: 22.99, H: 1.008, C: 12.01, O: 16.00).

  2. Write down the mole values for each element.

  3. Divide each mole value by the smallest number of moles to get the simplest ratio.

  4. Round to the nearest whole number to get the empirical formula subscripts.

  5. Stop here and set up the ratios for the student to finish.

Try solving on your own before revealing the answer!

Final Answer: NaHCO3

Moles: Na: ; H: ; C: ; O: .

Divide by 0.397: Na: 1, H: 1, C: 1, O: 3.

Empirical formula: NaHCO3

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