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Calculate the molar mass of the following compounds: CaCl2·2H2O, C6H12O6, Na2CO3, Fe2(SO4)3.

Chemistry

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1. (b) Calculate the molar mass of each of the following:
Given atomic masses: K=39\text{K}=39, Mn=55\text{Mn}=55, O=16\text{O}=16, H=1\text{H}=1, Ca=40\text{Ca}=40, Cl=35.5\text{Cl}=35.5, C=12\text{C}=12, Na=23\text{Na}=23, Fe=56\text{Fe}=56, N=14\text{N}=14.

  • (ii) CaCl2⋅2H2O\text{CaCl}_2 \cdot 2\text{H}_2\text{O} Molar mass=(1×40)+(2×35.5)+2×((2×1)+(1×16))\text{Molar mass} = (1 \times 40) + (2 \times 35.5) + 2 \times ((2 \times 1) + (1 \times 16)) =40+71+2×(2+16)= 40 + 71 + 2 \times (2 + 16) =111+2×18= 111 + 2 \times 18 =111+36=147 g/mol= 111 + 36 = \boxed{\text{147 g/mol}}
  • (iii) C6H12O6\text{C}_6\text{H}_{12}\text{O}_6 Molar mass=(6×12)+(12×1)+(6×16)\text{Molar mass} = (6 \times 12) + (12 \times 1) + (6 \times 16) =72+12+96=180 g/mol= 72 + 12 + 96 = \boxed{\text{180 g/mol}}
  • (iv) Na2CO3\text{Na}_2\text{CO}_3 Molar mass=(2×23)+(1×12)+(3×16)\text{Molar mass} = (2 \times 23) + (1 \times 12) + (3 \times 16) =46+12+48=106 g/mol= 46 + 12 + 48 = \boxed{\text{106 g/mol}}
  • (v) Fe2(SO4)3\text{Fe}_2(\text{SO}_4)_3 Molar mass=(2×56)+3×((1×32)+(4×16))\text{Molar mass} = (2 \times 56) + 3 \times ((1 \times 32) + (4 \times 16)) =112+3×(32+64)= 112 + 3 \times (32 + 64) =112+3×96= 112 + 3 \times 96 =112+288=400 g/mol= 112 + 288 = \boxed{\text{400 g/mol}}

2. (a) What is the molarity of sodium chloride weight in a solution prepared by dissolving 24.0g in enough water to make 250ml of solution (Na=23\text{Na}=23, Cl=35.5\text{Cl}=35.5)?

  1. Calculate the molar mass of NaCl\text{NaCl}.

    Molar mass of NaCl=23+35.5=58.5 g/mol\text{Molar mass of NaCl} = 23 + 35.5 = 58.5 \text{ g/mol}

  2. Calculate the number of moles of NaCl\text{NaCl}.

    Moles of NaCl=MassMolar mass=24.0 g58.5 g/mol≈0.4103 mol\text{Moles of NaCl} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{24.0 \text{ g}}{58.5 \text{ g/mol}} \approx 0.4103 \text{ mol}

  3. Convert the volume of the solution from milliliters to liters.

    Volume of solution=250 mL=0.250 L\text{Volume of solution} = 250 \text{ mL} = 0.250 \text{ L}

  4. Calculate the molarity of the solution.

    Molarity=Moles of soluteVolume of solution (L)=0.4103 mol0.250 L≈1.64 M\text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution (L)}} = \frac{0.4103 \text{ mol}}{0.250 \text{ L}} \approx \boxed{\text{1.64 M}}

    2. (b) The molecular mass of a compound is 104 g mol−1104 \text{ g mol}^{-1} and its empirical formula is CH\text{CH}. Deduce its molecular formula.

  5. Calculate the empirical formula mass of CH\text{CH}.

    Empirical formula mass of CH=(1×12)+(1×1)=13 g/mol\text{Empirical formula mass of CH} = (1 \times 12) + (1 \times 1) = 13 \text{ g/mol}

  6. Determine the ratio (n) between the molecular mass and the empirical formula mass.

    n=Molecular massEmpirical formula mass=104 g/mol13 g/mol=8n = \frac{\text{Molecular mass}}{\text{Empirical formula mass}} = \frac{104 \text{ g/mol}}{13 \text{ g/mol}} = 8

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

    Molecular formula=(CH)8=C8H8\text{Molecular formula} = (\text{CH})_8 = \boxed{\text{C}_8\text{H}_8}

    3. (a) N2(g)+3H2(g)⇌2NH3(g)\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightleftharpoons 2\text{NH}_3(\text{g})
    The above is a chemical equation

    • a. Read the equation defining all the symbols and notations This equation represents the reversible reaction where one mole of gaseous nitrogen (N2(g)\text{N}_2(\text{g})) reacts with three moles of gaseous hydrogen (3H2(g)3\text{H}_2(\text{g})) to produce two moles of gaseous ammonia (2NH3(g)2\text{NH}_3(\text{g})). The symbol (g)(\text{g}) indicates that the substances are in the gaseous state, and ⇌\rightleftharpoons signifies that the reaction is reversible, meaning it can proceed in both forward and reverse directions.

    • b. Explain the stoichiometry of the reaction. The stoichiometry of this reaction indicates the quantitative relationships between the reactants and products. Specifically, 1 mole of N2\text{N}_2 reacts with 3 moles of H2\text{H}_2 to produce 2 moles of NH3\text{NH}_3. This means that for every molecule of nitrogen, three molecules of hydrogen are consumed, yielding two molecules of ammonia. The mole ratio of N2:H2:NH3\text{N}_2 : \text{H}_2 : \text{NH}_3 is 1:3:21:3:2.

    • c. Explain the following:

      • (a) Limiting reactants: The limiting reactant (or limiting reagent) is the reactant in a chemical reaction that is completely consumed first, thereby determining the maximum amount of product that can be formed.
      • (b) Theoretical yield: The theoretical yield is the maximum amount of product that can be formed from a given amount of reactants, assuming the reaction goes to completion with 100% efficiency and no losses.
      • (c) Percentage yield: The percentage yield is a measure of the efficiency of a chemical reaction, calculated as the ratio of the actual yield (the amount of product experimentally obtained) to the theoretical yield (the maximum possible amount), multiplied by 100%. Percentage Yield=Actual YieldTheoretical Yield×100%\text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%

    3. (b) Calculate the theoretical yield (in grams) of potassium nitrite when 26.0g of potassium nitrate is heated with an excess of lead and the reaction Pb(s)+KNO3(s)→heatPbO+KNO2\text{Pb(s)} + \text{KNO}_3(\text{s}) \xrightarrow{\text{heat}} \text{PbO} + \text{KNO}_2 takes place.
    (Pb=207\text{Pb}=207, K=39\text{K}=39, N=14\text{N}=14, O=16\text{O}=16)

  8. Balance the chemical equation.

    The given equation is Pb(s)+KNO3(s)→heatPbO(s)+KNO2(s)\text{Pb(s)} + \text{KNO}_3(\text{s}) \xrightarrow{\text{heat}} \text{PbO(s)} + \text{KNO}_2(\text{s}).
    This equation is already balanced as written (1 Pb, 1 K, 1 N, 3 O on both sides).

  9. Calculate the molar mass of potassium nitrate (KNO3\text{KNO}_3).

    Molar mass of KNO3=(1×39)+(1×14)+(3×16)=39+14+48=101 g/mol\text{Molar mass of KNO}_3 = (1 \times 39) + (1 \times 14) + (3 \times 16) = 39 + 14 + 48 = 101 \text{ g/mol}

  10. Calculate the number of moles of KNO3\text{KNO}_3 from the given mass.

    Moles of KNO3=MassMolar mass=26.0 g101 g/mol≈0.2574 mol\text{Moles of KNO}_3 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{26.0 \text{ g}}{101 \text{ g/mol}} \approx 0.2574 \text{ mol}

  11. Use the mole ratio from the balanced equation to find the moles of potassium nitrite (KNO2\text{KNO}_2) produced.

    From the balanced equation, 1 mole of KNO3\text{KNO}_3 produces 1 mole of KNO2\text{KNO}_2.

    Moles of KNO2=Moles of KNO3=0.2574 mol\text{Moles of KNO}_2 = \text{Moles of KNO}_3 = 0.2574 \text{ mol}

  12. Calculate the molar mass of potassium nitrite (KNO2\text{KNO}_2).

    Molar mass of KNO2=(1×39)+(1×14)+(2×16)=39+14+32=85 g/mol\text{Molar mass of KNO}_2 = (1 \times 39) + (1 \times 14) + (2 \times 16) = 39 + 14 + 32 = 85 \text{ g/mol}

  13. Calculate the theoretical yield (mass) of KNO2\text{KNO}_2.

    Theoretical yield of KNO2=Moles×Molar mass=0.2574 mol×85 g/mol≈21.88 g\text{Theoretical yield of KNO}_2 = \text{Moles} \times \text{Molar mass} = 0.2574 \text{ mol} \times 85 \text{ g/mol} \approx \boxed{\text{21.88 g}}

    3. (c) If 16.0g of KNO2\text{KNO}_2 is produced, calculate the percentage yield.

  14. Use the actual yield (16.0 g) and the theoretical yield (21.88 g from 3b) to calculate the percentage yield.

    Percentage Yield=Actual YieldTheoretical Yield×100%\text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\% Percentage Yield=16.0 g21.88 g×100%≈73.12%\text{Percentage Yield} = \frac{16.0 \text{ g}}{21.88 \text{ g}} \times 100\% \approx \boxed{\text{73.12\%}}

    4. (a) Using a block diagram show the workings of a UV-VIS spectrophotometer.

    [Light Source] --> [Monochromator] --> [Sample Cell] --> [Detector] --> [Amplifier] --> [Readout/Display]
    • Light Source: Emits light across the UV and visible
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