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How many moles are present in 3.8 grams of lithium nitrate?

Sagot :

To determine the number of moles present in 3.8 grams of lithium nitrate (LiNO₃), we can follow these steps:

1. Identify the given mass of lithium nitrate: We are given that the mass of lithium nitrate is 3.8 grams.

2. Determine the molecular weight of lithium nitrate: This involves summing the atomic weights of each atom in the compound.
- Lithium (Li) has an atomic weight of 6.94.
- Nitrogen (N) has an atomic weight of 14.01.
- Each oxygen atom (O) has an atomic weight of 16.00, and there are three oxygen atoms in lithium nitrate.
Therefore, the molecular weight of LiNO₃ is calculated as follows:
[tex]\[ \text{Molecular weight of LiNO₃} = 6.94 + 14.01 + 3 \times 16.00 = 68.95 \][/tex]
Thus, the molecular weight of lithium nitrate is 68.95 grams per mole.

3. Calculate the number of moles: Using the formula:
[tex]\[ \text{Number of moles} = \frac{\text{mass of the substance}}{\text{molecular weight of the substance}} \][/tex]
We plug in the values:
[tex]\[ \text{Number of moles} = \frac{3.8 \, \text{grams}}{68.95 \, \text{grams per mole}} \][/tex]
Perform the division:
[tex]\[ \text{Number of moles} \approx 0.0551 \][/tex]

Hence, in 3.8 grams of lithium nitrate, there are approximately 0.0551 moles.