Answer :
Below is a step‐by‐step solution outlining each conversion.
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1. Convert moles to grams for each compound in part (1).
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(a) For 0.780 moles of calcium cyanide, [tex]$Ca(CN)_2$[/tex]:
1. Determine the molar mass. The atomic masses are:
• Calcium: 40.08 g/mol
• Carbon: 12.01 g/mol
• Nitrogen: 14.01 g/mol
There are two cyanide groups, and in each group the mass is
[tex]$$M(CN)=12.01+14.01=26.02\text{ g/mol}.$$[/tex]
Thus, the molar mass of [tex]$Ca(CN)_2$[/tex] is
[tex]$$M[Ca(CN)_2]=40.08+2(26.02)=40.08+52.04=92.12\text{ g/mol}.$$[/tex]
2. Calculate the mass:
[tex]$$\text{Mass}=0.780\text{ moles}\times 92.12\frac{\text{g}}{\text{mol}}=71.8536\text{ g}.$
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(b) For 2.50 moles of $\overline{F}(OH)_2$:
1. Determine the molar mass. Here, interpret the formula as containing one fluorine atom, two oxygen atoms, and two hydrogen atoms. The atomic masses are:
• Fluorine: 19.00 g/mol
• Oxygen: 16.00 g/mol
• Hydrogen: 1.008 g/mol
So, the molar mass is calculated as:
$$[/tex]M=19.00+2\times 16.00+2\times 1.008=19.00+32.00+2.016=53.016\text{ g/mol}.[tex]$$
2. Calculate the mass:
$$[/tex]\text{Mass}=2.50\text{ moles}\times 53.016\frac{\text{g}}{\text{mol}}=132.54\text{ g}.[tex]$
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(c) For 1.50 moles of pentane, $[/tex]C_5H_{12}[tex]$:
1. Calculate the molar mass. The atomic masses are:
• Carbon: 12.01 g/mol
• Hydrogen: 1.008 g/mol
Thus,
$[/tex][tex]$M(C_5H_{12})=5\times 12.01+12\times 1.008=60.05+12.096=72.146\text{ g/mol}.$[/tex][tex]$
2. Then,
$[/tex][tex]$\text{Mass}=1.50\text{ moles}\times 72.146\frac{\text{g}}{\text{mol}}=108.219\text{ g}.$[/tex]
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(d) For 0.00452 moles of [tex]$C_{20}H_{12}$[/tex]:
1. First, calculate the molar mass using:
• Carbon: 12.01 g/mol
• Hydrogen: 1.008 g/mol
So,
[tex]$$M(C_{20}H_{12})=20\times 12.01+12\times 1.008=240.20+12.096=252.296\text{ g/mol}.$$[/tex]
2. Then, compute the mass:
[tex]$$\text{Mass}=0.00452\text{ moles}\times 252.296\frac{\text{g}}{\text{mol}}=1.14037792\text{ g}.$
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2. Determine moles from a given mass in part (2).
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(9) For 3.54 g of hydrogen sulfide, $H_2S$:
1. Compute the molar mass. The atomic masses are:
• Hydrogen: 1.008 g/mol
• Sulfur: 32.06 g/mol
Thus,
$$[/tex]M(H_2S)=2\times 1.008+32.06=2.016+32.06=34.076\text{ g/mol}.[tex]$$
2. Calculate the number of moles:
$$[/tex]\text{Moles}=\frac{3.54\text{ g}}{34.076\frac{\text{g}}{\text{mol}}}\approx 0.10388543\text{ moles}.[tex]$
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For the remaining conversions in part (2), the formulas are not clearly defined, so no calculations can be performed.
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Final Answers:
1. Masses in grams:
• For $[/tex]C_{20}H_{12}[tex]$ (0.00452 moles): Molar mass = $[/tex][tex]$252.296\text{ g/mol}$[/tex][tex]$, Mass = $[/tex][tex]$1.14037792\text{ g}.$[/tex][tex]$
• For $[/tex]\overline{F}(OH)_2[tex]$ (2.50 moles): Molar mass = $[/tex][tex]$53.016\text{ g/mol}$[/tex][tex]$, Mass = $[/tex][tex]$132.54\text{ g}.$[/tex][tex]$
• For $[/tex]C_5H_{12}[tex]$ (1.50 moles): Molar mass = $[/tex][tex]$72.146\text{ g/mol}$[/tex][tex]$, Mass = $[/tex][tex]$108.219\text{ g}.$[/tex][tex]$
• For $[/tex]Ca(CN)_2[tex]$ (0.780 moles): Molar mass = $[/tex][tex]$92.12\text{ g/mol}$[/tex][tex]$, Mass = $[/tex][tex]$71.8536\text{ g}.$[/tex]
2. Moles from mass:
• For [tex]$H_2S$[/tex] (3.54 g): Molar mass = [tex]$$34.076\text{ g/mol}$$[/tex], Moles = [tex]$$[/tex]0.10388543\text{ moles}.$
These results give the complete solutions for the parts that are clearly defined.
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1. Convert moles to grams for each compound in part (1).
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(a) For 0.780 moles of calcium cyanide, [tex]$Ca(CN)_2$[/tex]:
1. Determine the molar mass. The atomic masses are:
• Calcium: 40.08 g/mol
• Carbon: 12.01 g/mol
• Nitrogen: 14.01 g/mol
There are two cyanide groups, and in each group the mass is
[tex]$$M(CN)=12.01+14.01=26.02\text{ g/mol}.$$[/tex]
Thus, the molar mass of [tex]$Ca(CN)_2$[/tex] is
[tex]$$M[Ca(CN)_2]=40.08+2(26.02)=40.08+52.04=92.12\text{ g/mol}.$$[/tex]
2. Calculate the mass:
[tex]$$\text{Mass}=0.780\text{ moles}\times 92.12\frac{\text{g}}{\text{mol}}=71.8536\text{ g}.$
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(b) For 2.50 moles of $\overline{F}(OH)_2$:
1. Determine the molar mass. Here, interpret the formula as containing one fluorine atom, two oxygen atoms, and two hydrogen atoms. The atomic masses are:
• Fluorine: 19.00 g/mol
• Oxygen: 16.00 g/mol
• Hydrogen: 1.008 g/mol
So, the molar mass is calculated as:
$$[/tex]M=19.00+2\times 16.00+2\times 1.008=19.00+32.00+2.016=53.016\text{ g/mol}.[tex]$$
2. Calculate the mass:
$$[/tex]\text{Mass}=2.50\text{ moles}\times 53.016\frac{\text{g}}{\text{mol}}=132.54\text{ g}.[tex]$
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(c) For 1.50 moles of pentane, $[/tex]C_5H_{12}[tex]$:
1. Calculate the molar mass. The atomic masses are:
• Carbon: 12.01 g/mol
• Hydrogen: 1.008 g/mol
Thus,
$[/tex][tex]$M(C_5H_{12})=5\times 12.01+12\times 1.008=60.05+12.096=72.146\text{ g/mol}.$[/tex][tex]$
2. Then,
$[/tex][tex]$\text{Mass}=1.50\text{ moles}\times 72.146\frac{\text{g}}{\text{mol}}=108.219\text{ g}.$[/tex]
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(d) For 0.00452 moles of [tex]$C_{20}H_{12}$[/tex]:
1. First, calculate the molar mass using:
• Carbon: 12.01 g/mol
• Hydrogen: 1.008 g/mol
So,
[tex]$$M(C_{20}H_{12})=20\times 12.01+12\times 1.008=240.20+12.096=252.296\text{ g/mol}.$$[/tex]
2. Then, compute the mass:
[tex]$$\text{Mass}=0.00452\text{ moles}\times 252.296\frac{\text{g}}{\text{mol}}=1.14037792\text{ g}.$
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2. Determine moles from a given mass in part (2).
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(9) For 3.54 g of hydrogen sulfide, $H_2S$:
1. Compute the molar mass. The atomic masses are:
• Hydrogen: 1.008 g/mol
• Sulfur: 32.06 g/mol
Thus,
$$[/tex]M(H_2S)=2\times 1.008+32.06=2.016+32.06=34.076\text{ g/mol}.[tex]$$
2. Calculate the number of moles:
$$[/tex]\text{Moles}=\frac{3.54\text{ g}}{34.076\frac{\text{g}}{\text{mol}}}\approx 0.10388543\text{ moles}.[tex]$
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For the remaining conversions in part (2), the formulas are not clearly defined, so no calculations can be performed.
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Final Answers:
1. Masses in grams:
• For $[/tex]C_{20}H_{12}[tex]$ (0.00452 moles): Molar mass = $[/tex][tex]$252.296\text{ g/mol}$[/tex][tex]$, Mass = $[/tex][tex]$1.14037792\text{ g}.$[/tex][tex]$
• For $[/tex]\overline{F}(OH)_2[tex]$ (2.50 moles): Molar mass = $[/tex][tex]$53.016\text{ g/mol}$[/tex][tex]$, Mass = $[/tex][tex]$132.54\text{ g}.$[/tex][tex]$
• For $[/tex]C_5H_{12}[tex]$ (1.50 moles): Molar mass = $[/tex][tex]$72.146\text{ g/mol}$[/tex][tex]$, Mass = $[/tex][tex]$108.219\text{ g}.$[/tex][tex]$
• For $[/tex]Ca(CN)_2[tex]$ (0.780 moles): Molar mass = $[/tex][tex]$92.12\text{ g/mol}$[/tex][tex]$, Mass = $[/tex][tex]$71.8536\text{ g}.$[/tex]
2. Moles from mass:
• For [tex]$H_2S$[/tex] (3.54 g): Molar mass = [tex]$$34.076\text{ g/mol}$$[/tex], Moles = [tex]$$[/tex]0.10388543\text{ moles}.$
These results give the complete solutions for the parts that are clearly defined.
