Question

final for semester 1
name
online portion (__ / 93 pts) + paper portion ( /13) = (__ / 106 pts)

1. molar mass: calculate the molar mass for the following compounds.

fe₂(so₄)₃
nh₃
baf₂

1. mole conversions - complete each conversion. molar masses calculated above may be used. (____ /10 pts)

a. how many moles are present in 153 grams of fe₂(so₄)₃? show work!
(____ / 3 pts)
answer:
b. how many molecules are present in 2.83 mol of nh₃? show work!
(____ / 3 pts)
answer:
c. how many grams are present in 3.4×10²⁴ cules of baf₂? show work!
(____ / 4 pts)
answer:

Explanation:

Part 1: Molar Mass Calculation

For $\boldsymbol{Fe_2(SO_4)_3}$

• Atomic masses: $Fe = 55.85\ \text{g/mol}$, $S = 32.07\ \text{g/mol}$, $O = 16.00\ \text{g/mol}$
• Molar mass $= 2(Fe) + 3(S) + 12(O)$

$= 2(55.85) + 3(32.07) + 12(16.00)$
$= 111.7 + 96.21 + 192$
$= 399.91\ \text{g/mol}$ (≈ $400\ \text{g/mol}$)

For $\boldsymbol{NH_3}$

• Atomic masses: $N = 14.01\ \text{g/mol}$, $H = 1.008\ \text{g/mol}$
• Molar mass $= N + 3(H)$

$= 14.01 + 3(1.008)$
$= 14.01 + 3.024$
$= 17.034\ \text{g/mol}$ (≈ $17.03\ \text{g/mol}$)

For $\boldsymbol{BaF_2}$

• Atomic masses: $Ba = 137.33\ \text{g/mol}$, $F = 19.00\ \text{g/mol}$
• Molar mass $= Ba + 2(F)$

$= 137.33 + 2(19.00)$
$= 137.33 + 38.00$
$= 175.33\ \text{g/mol}$

Part 2: Mole Conversions

2a. Moles in 153 g of $\boldsymbol{Fe_2(SO_4)_3}$

• Formula: $\text{Moles} = \frac{\text{Mass (g)}}{\text{Molar Mass (g/mol)}}$
• Substitute: $\text{Moles} = \frac{153\ \text{g}}{399.91\ \text{g/mol}} \approx 0.383\ \text{mol}$

2b. Molecules in 2.83 mol of $\boldsymbol{NH_3}$

• Avogadro’s number: $N_A = 6.022 \times 10^{23}\ \text{molecules/mol}$
• Formula: $\text{Molecules} = \text{Moles} \times N_A$
• Substitute: $\text{Molecules} = 2.83\ \text{mol} \times 6.022 \times 10^{23}\ \text{molecules/mol} \approx 1.70 \times 10^{24}\ \text{molecules}$

2c. Grams in $3.4 \times 10^{23}$ molecules of $\boldsymbol{BaF_2}$

• Step 1: Moles $= \frac{\text{Molecules}}{N_A} = \frac{3.4 \times 10^{23}}{6.022 \times 10^{23}\ \text{mol}^{-1}} \approx 0.5646\ \text{mol}$
• Step 2: Mass $= \text{Moles} \times \text{Molar Mass} = 0.5646\ \text{mol} \times 175.33\ \text{g/mol} \approx 98.9\ \text{g}$

Final Answers

1. Molar Masses:

• $Fe_2(SO_4)_3$: $\boldsymbol{399.91\ \text{g/mol}}$ (≈ $400\ \text{g/mol}$)
• $NH_3$: $\boldsymbol{17.03\ \text{g/mol}}$
• $BaF_2$: $\boldsymbol{175.33\ \text{g/mol}}$

1. Mole Conversions:

a. $\boldsymbol{0.383\ \text{mol}}$ (or ≈ $0.38\ \text{mol}$)
b. $\boldsymbol{1.70 \times 10^{24}\ \text{molecules}}$
c. $\boldsymbol{98.9\ \text{g}}$ (or ≈ $99\ \text{g}$)

Answer:

Part 1: Molar Mass Calculation

For $\boldsymbol{Fe_2(SO_4)_3}$

• Atomic masses: $Fe = 55.85\ \text{g/mol}$, $S = 32.07\ \text{g/mol}$, $O = 16.00\ \text{g/mol}$
• Molar mass $= 2(Fe) + 3(S) + 12(O)$

$= 2(55.85) + 3(32.07) + 12(16.00)$
$= 111.7 + 96.21 + 192$
$= 399.91\ \text{g/mol}$ (≈ $400\ \text{g/mol}$)

For $\boldsymbol{NH_3}$

• Atomic masses: $N = 14.01\ \text{g/mol}$, $H = 1.008\ \text{g/mol}$
• Molar mass $= N + 3(H)$

$= 14.01 + 3(1.008)$
$= 14.01 + 3.024$
$= 17.034\ \text{g/mol}$ (≈ $17.03\ \text{g/mol}$)

For $\boldsymbol{BaF_2}$

• Atomic masses: $Ba = 137.33\ \text{g/mol}$, $F = 19.00\ \text{g/mol}$
• Molar mass $= Ba + 2(F)$

$= 137.33 + 2(19.00)$
$= 137.33 + 38.00$
$= 175.33\ \text{g/mol}$

Part 2: Mole Conversions

2a. Moles in 153 g of $\boldsymbol{Fe_2(SO_4)_3}$

• Formula: $\text{Moles} = \frac{\text{Mass (g)}}{\text{Molar Mass (g/mol)}}$
• Substitute: $\text{Moles} = \frac{153\ \text{g}}{399.91\ \text{g/mol}} \approx 0.383\ \text{mol}$

2b. Molecules in 2.83 mol of $\boldsymbol{NH_3}$

• Avogadro’s number: $N_A = 6.022 \times 10^{23}\ \text{molecules/mol}$
• Formula: $\text{Molecules} = \text{Moles} \times N_A$
• Substitute: $\text{Molecules} = 2.83\ \text{mol} \times 6.022 \times 10^{23}\ \text{molecules/mol} \approx 1.70 \times 10^{24}\ \text{molecules}$

2c. Grams in $3.4 \times 10^{23}$ molecules of $\boldsymbol{BaF_2}$

• Step 1: Moles $= \frac{\text{Molecules}}{N_A} = \frac{3.4 \times 10^{23}}{6.022 \times 10^{23}\ \text{mol}^{-1}} \approx 0.5646\ \text{mol}$
• Step 2: Mass $= \text{Moles} \times \text{Molar Mass} = 0.5646\ \text{mol} \times 175.33\ \text{g/mol} \approx 98.9\ \text{g}$

Final Answers

1. Molar Masses:

• $Fe_2(SO_4)_3$: $\boldsymbol{399.91\ \text{g/mol}}$ (≈ $400\ \text{g/mol}$)
• $NH_3$: $\boldsymbol{17.03\ \text{g/mol}}$
• $BaF_2$: $\boldsymbol{175.33\ \text{g/mol}}$

1. Mole Conversions:

a. $\boldsymbol{0.383\ \text{mol}}$ (or ≈ $0.38\ \text{mol}$)
b. $\boldsymbol{1.70 \times 10^{24}\ \text{molecules}}$
c. $\boldsymbol{98.9\ \text{g}}$ (or ≈ $99\ \text{g}$)