Chapters 1-4 Chemistry Foundations Study Guide Page 1

CHEM 1406 Chapters 1-4 Study Guide

Section 2: Metric Prefixes, Conversions, & Density

Topic Review

Metric Prefix & Abbreviations Organizer

Prefix Meanings

• Kilo (k): 103 (1 kg = 1000 g)
• Centi (c): 10-2 (100 cm = 1 m)
• Milli (m): 10-3 (1000 mL = 1 L)
• Micro (μ): 10-6 (106 μg = 1 g)

Unit Types

• Length: Meter (m), Centimeter (cm)
• Mass: Gram (g), Kilogram (kg)
• Volume: Liter (L), Milliliter (mL)
• Density: g/mL or g/cm3

Key Equivalences

• 1 lb = 454 g
• 1 in. = 2.54 cm
• 1 oz = 28.3 g
• 1 mL = 1 cm3 (cc)

Worked Example Pairs: Dimensional Analysis & Density

Type 1: Clinical Dosage Calculation

Solved Example:

Order: 0.125 g of ampicillin. On hand: 250 mg / 5.0 mL. How many mL are needed?

1. Convert g to mg: 0.125 g × 1000 mg/g = 125 mg.
2. Use dosage factor: 125 mg × (5.0 mL / 250 mg).
3. Calculate: 625 / 250 = 2.5 mL (reported to 2 sig figs due to 5.0 mL).

Your Turn: (Practice Problem)

Order: 0.150 g of amoxicillin. On hand: 125 mg / 5.0 mL. How many mL are needed?

Write your calculations and final answer here:

Type 2: Density and Volume to Mass (kg)

Solved Example:

Find mass in kg of 2.00 L of solution with a density of 1.15 g/mL.

1. Convert volume L to mL: 2.00 L = 2000 mL.
2. Apply density: 2000 mL × 1.15 g/mL = 2300 g.
3. Convert g to kg: 2300 g / 1000 = 2.30 kg (reported to 3 sig figs: 2.30 kg).

Your Turn: (Practice Problem)

Find mass in kg of 1.50 L of solution with a density of 1.09 g/mL.

Write your calculations and final answer here:

Specific Gravity (SG) Equation

Specific gravity is a unitless value that compares the density of a substance to the density of pure water (1.00 g/mL).

SG = Density of Substance / Density of Water = Density of Substance / 1.00 g/mL (no units!)

Chapters 1-4 Chemistry Foundations Study Guide Page 2

CHEM 1406 Chapters 1-4 Study Guide

Section 3: Matter, States of Matter, & Energy

Topic Review

Physical vs. Chemical Changes

• Physical: Composition does not change. (e.g., boiling water, chopping wood, grinding beans, melting ice).
• Chemical: A new substance is formed with new properties. (e.g., iron rusting, coal burning, digestion, baking a cake).

Heating Curves & States of Matter

• Liquids & Gases: Both have randomly arranged particles, taking the shape of their container.
• Plateaus: Horizontal flat sections of a heating/cooling curve represent a change of state (constant temperature).

Worked Example Pairs: Specific Heat & Nutrition

Type 1: Specific Heat Energy Equation (q = m × c × ΔT)

Solved Example:

Calculate calories required to heat 35.0 g of iron from 25°C to 35°C. (c = 0.108 cal/g°C)

1. Identify inputs: m = 35.0 g, c = 0.108, ΔT = 35 - 25 = 10.°C (2 sig figs).
2. Multiply terms: q = 35.0 × 0.108 × 10. = 37.8 cal.
3. Round to 2 sig figs: 38 cal.

Your Turn: (Practice Problem)

Calculate calories required to heat 45.0 g of copper from 22°C to 37°C. (c = 0.0920 cal/g°C)

Write your calculations and final answer here:

Type 2: Caloric Values and Nutrition Bar Algebra

Solved Example:

A slice has 28 g carb, 13 g protein, and some fat. Total is 280 kcal. How many g of fat are present? (Carb/Protein = 4 kcal/g, Fat = 9 kcal/g)

1. Carb kcal: 28 × 4 = 112 → 110 kcal (tens place rounding).
2. Protein kcal: 13 × 4 = 52 → 50 kcal (rounded).
3. Fat kcal: 280 - (110 + 50) = 120 kcal.
4. Fat grams: 120 kcal / 9 kcal/g = 13.3 g → 13 g.

Your Turn: (Practice Problem)

A nutrition bar has 22 g carb, 8 g protein, and some fat. Total is 210 kcal. Find fat grams. (Caloric values same; round each food group to tens place!)

Write your calculations and final answer here:

Kelvin Temperature Conversion Rule

To convert Celsius to Kelvin, simply add 273.15 (rounded to 273 for whole degrees).

K = °C + 273

Chapters 1-4 Chemistry Foundations Study Guide Page 3

CHEM 1406 Chapters 1-4 Study Guide

Section 4: Atomic Structure & Periodic Trends

Topic Review

Subatomic Particles Summary

Worked Example Pairs: Atomic Symbols & dot notations

Type 1: Finding Subatomic Count from Symbol AZX

Solved Example:

Find p+, n0, e- in neutral 20882Pb.

1. Atomic number (Z = 82): represents 82 protons.
2. In a neutral atom, protons = electrons, so 82 electrons.
3. Mass number (A = 208) = protons + neutrons.
4. Neutrons = 208 - 82 = 126 neutrons.

Your Turn: (Practice Problem)

Find p+, n0, e- in neutral 19779Au.

Write your protons, neutrons, and electrons counts here:

Type 2: Valence Electrons and Lewis Symbols

Solved Example:

Draw electron dot symbol for Carbon.

• Carbon is in Group 4A (14).
• It has 4 valence electrons.
• Distribute electrons as single dots on each of the four sides first. No pairs!
• Result: 4 single dots around the C.

Your Turn: (Practice Problem)

Draw electron dot symbol for Nitrogen.

Determine its group and valence count, then draw the dot arrangement around 'N':

Periodic Table Landmarks Reminder

Group 1A (Alkali Metals): Highly reactive, shiny solids; good heat & electricity conductors.

Group 7A (Halogens): Highly reactive diatomic nonmetals; very poor conductors.

Energy Levels (Shells): Equal to the period number (sodium in Period 3 has 3 levels).

Chapters 1-4 Chemistry Foundations Study Guide Page 4

• Q11 (E): 4.080 × 10-5 has 4 sig figs. The trapped zero and the trailing zero in a decimal number are both significant.
• Q12 (B): Rounding 537.8241 to three sig figs means looking at the tenths digit (8), which rounds 7 up to 8. Report 538.
• Q13 (D): Denominator is 18 × 4.02 = 72.36. Since 18 has 2 sig figs, denominator is 72. Then 36.24 / 72.36 = 0.5008..., which rounds to 2 sig figs: 0.50 mL.
• Q14 (E): Sum: 3520 + 14.8 + 0.35 = 3535.15. Since 3520 is precise only to the tens place, the final sum rounds to the tens place: 3540.

Q17, Q19, Q20: Dimensional Analysis & Dosages

• Q17 (C): 85 g × (1 oz / 28.3 g) = 3.0035 oz. Rounding to 2 sig figs (limited by 85 g) gives 3.0 oz.
• Q19 (D): 4.5 m × (100 cm / 1 m) × (1 in. / 2.54 cm) = 177.16 in.. Rounded to 2 sig figs (from 4.5 m) gives 180 in.
• Q20 (D): Order is 0.150 g. Convert to mg: 0.150 g × 1000 mg/g = 150 mg. Apply concentration: 150 mg × (5.0 mL / 125 mg) = 6.0 mL.

Q21, Q22: Density & Specific Gravity

• Q21 (A): 1.50 L = 1500 mL. Mass: 1500 mL × 1.09 g/mL = 1635 g. Convert to kg: 1635 g / 1000 = 1.635 kg. Round to 3 sig figs: 1.64 kg.
• Q22 (B): Density is 41.2 g / 40.0 mL = 1.03 g/mL. Specific gravity: 1.03 g/mL / 1.00 g/mL = 1.03 (unitless!).

CHEM 1406 Practice Exam Solutions Key Page 1

CHEM 1406 Practice Exam Solutions Key (Chapters 1-4) Practice Exam

Step-by-Step Worked Calculations (Q23 - Q40)

Q27, Q28: Kelvin and Energy Units Conversions

• Q27 (B): To find Kelvin: K = °C + 273.15. For -40 °C, we calculate -40 + 273 = 233 K.
• Q28 (C): Convert Joules to calories using the equivalence factor 1 cal = 4.184 J. Calculation: 820 J × (1 cal / 4.184 J) = 195.98 cal. Rounded to 3 sig figs: 196 cal.

Q29: Specific Heat Heat Transfer Math

Find calories required to heat a copper sample: q = m × c × ΔT

1. Determine change in temperature: ΔT = Tfinal - Tinitial = 37 °C - 22 °C = 15 °C (2 significant figures).
2. Plug values into equation: q = 45.0 g × 0.0920 cal/g°C × 15 °C.
3. Calculate product: q = 62.1 cal.
4. Round to 2 sig figs (due to ΔT limits): 62 cal (Choice E).

Q31: Food Energy Algebra

Calculate fat content from total kcal:

1. Determine carbohydrate energy: 22 g × 4 kcal/g = 88 kcal → 90 kcal (rounded to the tens place).
2. Determine protein energy: 8 g × 4 kcal/g = 32 kcal → 30 kcal (rounded to the tens place).
3. Find remaining energy for fat: 210 kcal - (90 + 30) = 210 - 120 = 90 kcal.
4. Convert fat energy to grams: 90 kcal / (9 kcal/g) = 10 g of fat (Choice E).

Q37: Isotope Subatomic Counts

For isotopic symbol 19779Au:

• Bottom number (Z) is atomic number = 79 protons.
• Neutral atom requires protons = electrons = 79 electrons.
• Top number (A) is mass number = protons + neutrons.
• Subtract atomic number from mass number: 197 - 79 = 118 neutrons (Choice C).

Q38, Q39, Q40: Electron Arrangements

• Q38 (C): Sodium (Na) resides in Period 3 of the periodic table, so it possesses 3 electron energy shells.
• Q39 (D): For Group A elements (main-group elements), the number of valence electrons is always equal to its Group A number (e.g., Group 1A elements have 1 valence e-).
• Q40 (C): Nitrogen (N) is in Group 5A, meaning it possesses 5 valence electrons. The correct dot representation displays one pair of dots and three individual single dots distributed around the symbol (Choice C).

CHEM 1406 Practice Exam Solutions Key Page 2