Week 2 Practice Quiz — Answers & Walkthrough

What You'll Practice

1Molar Mass of CaCO₃
2Counting Oxygen Atoms in Glucose
3Molecules to Moles Conversion
4Balancing a Chemical Equation
5Classifying Reaction Types
6Endothermic vs. Exothermic
7Stoichiometry — Mole to Mole
8Stoichiometry — Mole to Gram
9Ideal Gas Law (PV = nRT)
10Dalton's Law of Partial Pressures

Molar Mass

Calculate the Molar Mass of CaCO₃

Calculate the molar mass of calcium carbonate (CaCO₃).

Try It — Enter your answer (g/mol):

Step 1 — Identify Each Element

Break the formula into its atoms: Ca, C, and O (×3).

Step 2 — Look Up Atomic Masses

Atom	Count	Atomic Mass	Subtotal
Ca	1	40.08	40.08
C	1	12.01	12.01
O	3	16.00	48.00

Step 3 — Add Them Up

40.08 + 12.01 + 48.00 = 100.09 g/mol

Final Answer

100.09 g/mol

Study Tip: Always multiply atomic mass × number of atoms for each element, then sum everything. This is the foundation for every stoichiometry problem.

Mole Ratios in a Formula

Moles of Oxygen in Glucose

A student has 1.5 moles of glucose (C₆H₁₂O₆). How many moles of oxygen atoms are in the sample?

Try It — Enter your answer (mol O):

Step 1 — Read the Subscript

The formula C₆H₁₂O₆ tells us each molecule has 6 oxygen atoms.

Step 2 — Multiply

1.5 mol glucose × 6 mol O / 1 mol glucose

Step 3 — Solve

= 9.0 mol O atoms

Final Answer

9.0 moles of oxygen atoms

Study Tip: Subscripts in a formula directly give you the mole-to-mole ratio of atoms within one molecule. This concept extends to every formula you encounter.

Avogadro's Number

Molecules to Moles

Convert 3.01 × 10²³ molecules of water (H₂O) into moles.

Try It — Enter your answer (mol):

Step 1 — Recall Avogadro's Number

1 mole = 6.022 × 10²³ particles (molecules, atoms, etc.)

Step 2 — Set Up the Conversion

moles = molecules ÷ 6.022 × 10²³

Step 3 — Calculate

3.01 × 10²³ ÷ 6.022 × 10²³ = 0.500 mol

Final Answer

0.500 mol H₂O

Study Tip: Think of Avogadro's number as the "dozen" of chemistry. Just as 12 items = 1 dozen, 6.022 × 10²³ particles = 1 mole.

Balancing Equations

Al + O₂ → Al₂O₃

Balance the equation: __Al + __O₂ → __Al₂O₃

Try It — Enter coefficients (e.g. "4,3,2"):

Step 1 — Count Atoms (Unbalanced)

Element	Left	Right
Al	1	2
O	2	3

Step 2 — Balance Oxygen First

LCM of 2 and 3 = 6. Place 3 in front of O₂ (gives 6 O on left) and 2 in front of Al₂O₃ (gives 6 O on right).

Step 3 — Balance Aluminum

Right side now has 2 × 2 = 4 Al. Place 4 in front of Al on the left.

4Al + 3O₂ → 2Al₂O₃

Step 4 — Verify

Element	Left	Right
Al	4	4
O	6	6

Balanced!

Final Answer

4Al + 3O₂ → 2Al₂O₃

Study Tip: Start by balancing the element that appears in the fewest formulas, then work outward. Always verify your count at the end.

Reaction Types

Classify 2KClO₃ → 2KCl + 3O₂

Classify this reaction: 2KClO₃ → 2KCl + 3O₂
(Synthesis, Decomposition, Single Replacement, or Double Replacement?)

Try It — Type your answer:

Step 1 — Count Reactants vs. Products

1 reactant (KClO₃) → 2 products (KCl + O₂)

Step 2 — Match the Pattern

Type	Pattern
Synthesis	A + B → AB
Decomposition	AB → A + B
Single Replacement	A + BC → AC + B
Double Replacement	AB + CD → AD + CB

One compound breaks into two or more products = Decomposition.

Final Answer

Decomposition

Study Tip: "De-compose" = to break apart. If one thing becomes many, it's decomposition. If many become one, it's synthesis.

Thermochemistry

Endothermic or Exothermic?

A student dissolves ammonium nitrate (NH₄NO₃) in water and the beaker becomes very cold to the touch. Is this endothermic or exothermic?

Try It — Type your answer:

Step 1 — Observe the Evidence

The beaker feels cold. This means heat is leaving your hand and entering the solution — the solution is absorbing energy from the surroundings.

Step 2 — Apply the Definitions

Process	Energy Flow	Surroundings Feel…
Endothermic	Absorbs heat	Cold
Exothermic	Releases heat	Hot

Final Answer

Endothermic

Study Tip: "Endo" = in (energy goes into the system). Think of instant cold packs — they use NH₄NO₃ dissolving in water, an endothermic process.

Stoichiometry

Mole-to-Mole Conversion

Using: 2H₂ + O₂ → 2H₂O
If you start with 4.0 moles of H₂, how many moles of H₂O are produced?

Try It — Enter your answer (mol H₂O):

Step 1 — Identify the Mole Ratio

From the balanced equation: 2 mol H₂ : 2 mol H₂O
This simplifies to a 1 : 1 ratio.

Step 2 — Apply the Ratio

4.0 mol H₂ × (2 mol H₂O / 2 mol H₂) = 4.0 mol H₂O

Final Answer

4.0 mol H₂O

Study Tip: Coefficients in a balanced equation are your mole ratios. Always start stoichiometry by writing the ratio between the substance you have and the substance you want.

Stoichiometry

Mole-to-Gram Conversion

Using: 2H₂ + O₂ → 2H₂O
How many grams of H₂O are produced from 2.0 moles of O₂?

Try It — Enter your answer (grams):

Step 1 — Mole Ratio

From the equation: 1 mol O₂ : 2 mol H₂O

2.0 mol O₂ × (2 mol H₂O / 1 mol O₂) = 4.0 mol H₂O

Step 2 — Convert Moles to Grams

Molar mass of H₂O = 2(1.01) + 16.00 = 18.02 g/mol

4.0 mol × 18.02 g/mol = 72.08 g

Final Answer

72.08 g H₂O

Study Tip: The stoichiometry roadmap: Grams → Moles → (use ratio) → Moles → Grams. Always convert to moles first before crossing the equation bridge.

Gas Laws

Ideal Gas Law (PV = nRT)

A 3.5 mol sample of N₂ gas is in a 10.0 L container at 37.0°C. Calculate the pressure.

Try It — Enter your answer (atm):

Step 1 — Convert Temperature to Kelvin

T = 37.0 + 273.15 = 310.15 K

Gas law calculations always require Kelvin!

Step 2 — List Known Values

Variable	Value
n	3.5 mol
V	10.0 L
T	310.15 K
R	0.08206 L·atm/mol·K

Step 3 — Solve for P

P = nRT / V

P = (3.5 × 0.08206 × 310.15) / 10.0

P = 89.04 / 10.0 = 8.9 atm

Final Answer

P ≈ 8.9 atm

Study Tip: The #1 mistake on gas law problems is forgetting to convert °C to K. Make it your first step every time. Also remember which R value matches your units.

Gas Laws

Dalton's Law of Partial Pressures

A mixture of three gases has a total pressure of 2.50 atm. Gas A = 0.80 atm, Gas B = 1.05 atm. What is the partial pressure of Gas C?

Try It — Enter your answer (atm):

Step 1 — State Dalton's Law

P_total = P_A + P_B + P_C

The total pressure equals the sum of all individual (partial) pressures.

Step 2 — Rearrange and Solve

P_C = P_total − P_A − P_B

P_C = 2.50 − 0.80 − 1.05 = 0.65 atm

Final Answer

P_C = 0.65 atm

Study Tip: Dalton's Law is straightforward addition/subtraction. In clinical settings, this is how we understand O₂ and CO₂ partial pressures in blood gas analysis (ABGs).

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Review Complete!

You've worked through all 10 chemistry problems. Revisit any question from the table of contents to reinforce the concepts.