How many grams of lithium bromide are present in a 15.50 mole sample?
2. How many moles of copper (II) cyanide are present in a 3.40 x 10 cubed mg sample?
3. How many moles of cobalt (1) carbonate are present in a sample containing 1.75 g of cobalt?
The Correct Answer and Explanation is :
Let’s break down and solve each of these questions step-by-step.
Question 1: How many grams of lithium bromide are present in a 15.50 mole sample?
We are asked to find the mass of lithium bromide (LiBr) in grams.
Step 1: Use the molar mass of LiBr.
The molar mass of lithium bromide is calculated by adding the atomic masses of lithium (Li) and bromine (Br):
- Lithium (Li) has an atomic mass of about 6.94 g/mol.
- Bromine (Br) has an atomic mass of about 79.90 g/mol.
So, the molar mass of LiBr = 6.94 g/mol + 79.90 g/mol = 86.84 g/mol.
Step 2: Apply the formula:
[ \text{Mass (g)} = \text{Moles (mol)} \times \text{Molar Mass (g/mol)} ]
[ \text{Mass} = 15.50 \, \text{mol} \times 86.84 \, \text{g/mol} ]
[ \text{Mass} = 1346.04 \, \text{g} ]
Answer: There are 1346.04 grams of lithium bromide in a 15.50 mole sample.
Question 2: How many moles of copper (II) cyanide are present in a 3.40 x 10³ mg sample?
We are asked to find the number of moles of copper (II) cyanide (Cu(CN)₂) in a given mass of 3.40 x 10³ milligrams (mg).
Step 1: Convert milligrams to grams.
1 gram = 1000 milligrams, so:
[ 3.40 \times 10^3 \, \text{mg} = \frac{3.40 \times 10^3}{1000} = 3.40 \, \text{g} ]
Step 2: Find the molar mass of copper (II) cyanide.
The molar mass of Cu(CN)₂ is calculated by adding the atomic masses of copper (Cu), carbon (C), and nitrogen (N):
- Copper (Cu) = 63.55 g/mol
- Carbon (C) = 12.01 g/mol
- Nitrogen (N) = 14.01 g/mol
So, the molar mass of Cu(CN)₂:
[ 63.55 \, \text{g/mol} + (2 \times (12.01 \, \text{g/mol} + 14.01 \, \text{g/mol})) ]
[ = 63.55 \, \text{g/mol} + (2 \times 26.02 \, \text{g/mol}) ]
[ = 63.55 \, \text{g/mol} + 52.04 \, \text{g/mol} ]
[ = 115.59 \, \text{g/mol} ]
Step 3: Use the formula to find moles.
[
\text{Moles} = \frac{\text{Mass (g)}}{\text{Molar Mass (g/mol)}}
]
[
\text{Moles} = \frac{3.40 \, \text{g}}{115.59 \, \text{g/mol}} \approx 0.0294 \, \text{mol}
]
Answer: There are 0.0294 moles of copper (II) cyanide in the 3.40 x 10³ mg sample.
Question 3: How many moles of cobalt (I) carbonate are present in a sample containing 1.75 g of cobalt?
We are asked to find the number of moles of cobalt (I) carbonate (Co₂CO₃) from the given cobalt mass.
Step 1: Find the molar mass of cobalt (I) carbonate.
Cobalt (I) carbonate has the formula Co₂CO₃, where:
- Cobalt (Co) has an atomic mass of 58.93 g/mol.
- Carbon (C) has an atomic mass of 12.01 g/mol.
- Oxygen (O) has an atomic mass of 16.00 g/mol.
The molar mass of Co₂CO₃:
[
(2 \times 58.93 \, \text{g/mol}) + 12.01 \, \text{g/mol} + (3 \times 16.00 \, \text{g/mol})
]
[
= 117.86 \, \text{g/mol} + 12.01 \, \text{g/mol} + 48.00 \, \text{g/mol}
]
[
= 177.87 \, \text{g/mol}
]
Step 2: Use the formula to find moles of cobalt.
We know that cobalt (I) carbonate contains two moles of cobalt per formula unit. First, we need to find the moles of cobalt (Co) in the sample.
[
\text{Moles of Co} = \frac{\text{Mass of Co (g)}}{\text{Molar Mass of Co (g/mol)}}
]
[
\text{Moles of Co} = \frac{1.75 \, \text{g}}{58.93 \, \text{g/mol}} \approx 0.0297 \, \text{mol}
]
Step 3: Find the moles of cobalt (I) carbonate.
Since each mole of Co₂CO₃ contains 2 moles of cobalt, the number of moles of cobalt (I) carbonate is:
[
\text{Moles of Co₂CO₃} = \frac{\text{Moles of Co}}{2} = \frac{0.0297 \, \text{mol}}{2} \approx 0.01485 \, \text{mol}
]
Answer: There are 0.01485 moles of cobalt (I) carbonate in the sample containing 1.75 g of cobalt.
Summary:
- 15.50 moles of lithium bromide corresponds to 1346.04 grams.
- 3.40 x 10³ mg of copper (II) cyanide corresponds to 0.0294 moles.
- 1.75 g of cobalt corresponds to 0.01485 moles of cobalt (I) carbonate.