Moles Notes

Analysis of Hydrates in Chemistry

Introduction to Hydrates

  • Hydrates are compounds that contain water molecules within their crystalline structure.

  • The general formula for a hydrate can be expressed as:

    • MximesnH2OMx imes nH_2O

    • Here, M represents the metal salt, x is the number of moles of the salt, n is the number of moles of water.

Example Problem: Determining the Formula of a Hydrate

Given

  • Sample Information: Consider a hydrate of Calcium Sulfate, CaSO4CaSO_4.

  • Mass of Crucible Empty: mcrucible=1.002extgm_{crucible} = 1.002 ext{ g}

  • Mass of Crucible with Hydrate: mcrucible+hydrate=2.515extgm_{crucible+hydrate} = 2.515 ext{ g}

  • Mass After Heating (All Water Driven Off): To be determined.

Procedure

  1. Initial Mass Calculation:

    • Calculate the mass of the hydrate:
      mhydrate=mcrucible+hydrate−mcruciblem_{hydrate} = m_{crucible+hydrate} - m_{crucible}

    • Substitute values:
      mhydrate=2.515extg−1.002extg=1.513extgm_{hydrate} = 2.515 ext{ g} - 1.002 ext{ g} = 1.513 ext{ g}

  2. Final Mass Calculation:

    • The mass of the crucible after heating (mass after all water has been driven off) is calculated, labelled as mfinalm_{final}.

    • (This needs to be either provided in further detail, or determined.)

  3. Mass of Water Lost Calculation:

    • Determine the mass of water lost during the heating process:
      mwater=mhydrate−mfinalm_{water} = m_{hydrate} - m_{final}

  4. Mole Calculations:

    • Calculate moles of the anhydrous compound:
      nanhydrous=racmfinalMCaSO4n_{anhydrous} = rac{m_{final}}{M_{CaSO_4}}

    • Molar Mass of CaSO4CaSO_4 is approximately 136.14 g/mol.

  5. Moles of Water Calculation:

    • Calculate moles of water that were lost:
      nwater=racmwaterMH2On_{water} = rac{m_{water}}{M_{H_2O}}

    • Molar mass of water H2OH_2O is approximately 18.02 g/mol.

  6. Determine Hydrate Formula:

    • Use the ratio of moles of CaSO4CaSO_4 to moles of H2OH_2O to establish the formula:

    • For example, if you find there are 1 mole of CaSO4CaSO_4 and 2 moles of water, the formula of the hydrate will be CaSO4imes2H2OCaSO_4 imes 2H_2O.

Conclusion

  • This methodology allows one to systematically determine the formula of a hydrate using mass measurements before and after heating, thus facilitating the understanding of hydrous compounds in chemistry.

  • PART A – THE MOLE: BIG IDEAS (Read this first)

    What is a mole?

    A mole is just a counting number – like a dozen, but much bigger.

    · 1 dozen = 12 things

    · 1 mole = 6.02 \times 10^{23} things

    That number is called Avogadro’s number.

    We use moles because atoms and molecules are so tiny that counting them one by one is impossible.

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    The 3 main mole relationships (memorize these)

    You have… To get moles…

    Mass (grams) Divide by molar mass

    Number of particles (molecules/atoms) Divide by 6.02 \times 10^{23}

    Volume of a gas at STP (Liters) Divide by 22.4 L/mol

    And backwards:

    · Moles → grams: multiply by molar mass

    · Moles → particles: multiply by 6.02 \times 10^{23}

    · Moles → gas volume at STP: multiply by 22.4 L/mol

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    PART B – MOLAR MASS (g/mol)

    How to calculate molar mass

    Add up the atomic masses from the periodic table for all atoms in the formula.

    Example: H₂O

    · H: 2 × 1.01 = 2.02

    · O: 1 × 16.00 = 16.00

    · Total = 18.02 g/mol

    Example: Ca(OH)₂

    · Ca: 1 × 40.08 = 40.08

    · O: 2 × 16.00 = 32.00

    · H: 2 × 1.01 = 2.02

    · Total = 74.10 g/mol

    Example with polyatomic ion: Al₂(SO₄)₃

    · Al: 2 × 26.98 = 53.96

    · S: 3 × 32.06 = 96.18

    · O: 12 × 16.00 = 192.00

    · Total = 342.14 g/mol

    Hydrate example: Na₂SO₄ · 10H₂O

    · Na₂SO₄ = 142.04 g/mol

    · 10 H₂O = 10 × 18.02 = 180.20 g/mol

    · Total = 322.24 g/mol

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    PART C – MOLE CONVERSIONS (The 3 paths)

    Path 1: Mass ↔ Moles

    \text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}

    \text{mass} = \text{moles} \times \text{molar mass}

    Example: How many moles in 10.0 g of NaCl?

    Molar mass NaCl = 58.44 g/mol

    n = 10.0 / 58.44 = 0.171 mol

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    Path 2: Particles ↔ Moles

    \text{moles} = \frac{\text{number of particles}}{6.02 \times 10^{23}}

    \text{particles} = \text{moles} \times 6.02 \times 10^{23}

    Example: How many molecules in 2.00 mol of H₂O?

    2.00 \times 6.02 \times 10^{23} = 1.20 \times 10^{24} molecules

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    Path 3: Gas Volume at STP ↔ Moles

    At STP (0°C, 101.3 kPa): 1 mol gas = 22.4 L

    \text{moles} = \frac{\text{volume (L)}}{22.4}

    \text{volume (L)} = \text{moles} \times 22.4

    Example: Volume of 0.350 mol SO₂ at STP?

    0.350 \times 22.4 = 7.84 L

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    PART D – TWO-STEP PROBLEMS (Particles → Mass, etc.)

    Example: How many molecules in 345 g of NaOH?

    Step 1 – moles: 345 / 40.00 = 8.625 mol

    Step 2 – molecules: 8.625 \times 6.02 \times 10^{23} = 5.19 \times 10^{24} molecules

    Example: Mass of 2.68 \times 10^{21} molecules of H₂O?

    Step 1 – moles: (2.68 \times 10^{21}) / (6.02 \times 10^{23}) = 0.00445 mol

    Step 2 – mass: 0.00445 \times 18.02 = 0.0802 g

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    PART E – PERCENT COMPOSITION

    Formula:

    \% \text{ element} = \frac{\text{mass of element in 1 mole of compound}}{\text{molar mass of compound}} \times 100\%

    Example: % Al in AlCl₃

    · Molar mass AlCl₃ = 26.98 + (3 × 35.45) = 133.33 g/mol

    · Mass of Al = 26.98 g

    · \% = (26.98 / 133.33) \times 100 = 20.24\%

    Check: % Cl = (106.35 / 133.33) \times 100 = 79.76\%

    Total = 100%

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    PART F – EMPIRICAL & MOLECULAR FORMULAS

    Definitions

    · Empirical formula: smallest whole number ratio of atoms (e.g., CH₂O)

    · Molecular formula: actual number of atoms in a molecule (e.g., C₆H₁₂O₆)

    Finding molecular formula (given empirical formula & molar mass)

    Formula:

    n = \frac{\text{molar mass of compound}}{\text{molar mass of empirical formula}}

    Then: molecular formula = (empirical formula) × n

    Example: Empirical formula = CH₂, molar mass = 70.0 g/mol

    · Molar mass CH₂ = 12.01 + (2 × 1.01) = 14.03 g/mol

    · n = 70.0 / 14.03 \approx 5

    · Molecular formula = C₅H₁₀

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    PART G – QUICK REFERENCE CARD

    Conversion Formula

    Mass → moles n = \frac{m}{MM}

    Moles → mass m = n \times MM

    Particles → moles n = \frac{\text{#particles}}{6.02 \times 10^{23}}

    Moles → particles \text{#particles} = n \times 6.02 \times 10^{23}

    Gas volume (L) at STP → moles n = \frac{V}{22.4}

    Moles → gas volume at STP V = n \times 22.4

    Percent composition \% = \frac{\text{element mass in 1 mol}}{MM} \times 100

    Empirical → molecular n = \frac{MM_{\text{compound}}}{MM_{\text{empirical}}}

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    PART H – TRICK QUESTIONS & HOW TO BEAT THEM

    Trick Fix

    Forgetting units on Avogadro’s number Always write 6.02 \times 10^{23} molecules/mol or atoms/mol

    Using 22.4 L/mol for non-STP Only works at STP (0°C, 101.3 kPa)

    Mixing up atoms vs molecules CO₂ molecule has 3 atoms; be careful what the question asks for

    Hydrates: forgetting the water Include the water molecules in molar mass

    Empirical formula not reducing Divide subscripts by GCD

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    PART I – SIG FIGS FOR MOLE PROBLEMS

    · Multiplication/division: fewest sig figs in given numbers

    · Molar mass from periodic table: use 2 decimal places (e.g., 16.00, 1.01, 12.01)

    Example: 10.0 g (3 sig figs) / 58.44 g/mol (4 sig figs) = 0.171 mol (3 sig figs)

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    PRACTICE PROBLEMS (Try these)

    1. Molar mass of Ca₃(PO₄)₂?

    2. Moles in 25.0 g of CO₂?

    3. Mass of 0.500 mol of glucose (C₆H₁₂O₆)?

    4. Molecules in 5.00 g of H₂O?

    5. Volume at STP of 2.00 mol of N₂?

    6. % oxygen in H₂SO₄?

    7. Empirical formula CH₂, molar mass 84 g/mol. Molecular formula?

    Answers:

    1. 310.18 g/mol

    2. 0.568 mol

    3. 90.08 g

    4. 1.67 \times 10^{23} molecules

    5. 44.8 L

    6. 65.25%

    7. C₆H₁₂