. What are the masses of 3-methyl-2-butanone and 2,3-dimethyl-3-pentanol that are
present in the sample if the sample weighs 5.13 g and contains only those two
compounds?
To determine the mass of each compound present in the sample, we need to know the mole ratio between them.
The chemical formula for 3-methyl-2-butanone is C5H10O, and the molar mass can be calculated as:
(3 x 12.01 g/mol) + (7 x 1.01 g/mol) + (1 x 16.00 g/mol) = 86.13 g/mol
The chemical formula for 2,3-dimethyl-3-pentanol is C7H16O, and the molar mass can be calculated as:
(7 x 12.01 g/mol) + (16 x 1.01 g/mol) + (1 x 16.00 g/mol) = 130.23 g/mol
Now, we can calculate the moles of each compound in the 5.13 g sample:
Mass of 3-methyl-2-butanone = (mass of sample) x (mol ratio of 3-methyl-2-butanone / total mol ratio)
Mass of 3-methyl-2-butanone = 5.13 g x (86.13 g/mol / (86.13 g/mol + 130.23 g/mol))
Mass of 3-methyl-2-butanone = 5.13 g x (86.13 / 216.36) = 2.04 g
Mass of 2,3-dimethyl-3-pentanol = (mass of sample) x (mol ratio of 2,3-dimethyl-3-pentanol / total mol ratio)
Mass of 2,3-dimethyl-3-pentanol = 5.13 g x (130.23 g/mol / (86.13 g/mol + 130.23 g/mol))
Mass of 2,3-dimethyl-3-pentanol = 5.13 g x (130.23 / 216.36) = 3.09 g
Therefore, there are approximately 2.04 g of 3-methyl-2-butanone and 3.09 g of 2,3-dimethyl-3-pentanol present in the sample.
To determine the masses of 3-methyl-2-butanone and 2,3-dimethyl-3-pentanol present in a sample, you will need to use the concept of percent composition. The percent composition expresses the mass of each component in a compound as a percentage of the total mass of the compound.
Step 1: Find the percent composition of each compound.
- The molar mass of 3-methyl-2-butanone is calculated as follows:
- Carbon (C): 3 atoms x 12.01 g/mol = 36.03 g/mol
- Hydrogen (H): 6 atoms x 1.01 g/mol = 6.06 g/mol
- Oxygen (O): 1 atom x 16.00 g/mol = 16.00 g/mol
- Total: 36.03 g/mol + 6.06 g/mol + 16.00 g/mol = 58.09 g/mol
- The molar mass of 2,3-dimethyl-3-pentanol is calculated as follows:
- Carbon (C): 7 atoms x 12.01 g/mol = 84.07 g/mol
- Hydrogen (H): 16 atoms x 1.01 g/mol = 16.16 g/mol
- Oxygen (O): 1 atom x 16.00 g/mol = 16.00 g/mol
- Total: 84.07 g/mol + 16.16 g/mol + 16.00 g/mol = 116.23 g/mol
Step 2: Calculate the percent composition.
- To find the percent composition of each compound, divide its molar mass by the total molar mass of the sample, and then multiply by 100.
- For 3-methyl-2-butanone:
- Percent composition = (molar mass of 3-methyl-2-butanone / total molar mass) x 100
- Percent composition = (58.09 g/mol / (58.09 g/mol + 116.23 g/mol)) x 100
- Percent composition = (58.09 g/mol / 174.32 g/mol) x 100
- Percent composition = 0.3333 x 100
- Percent composition = 33.33%
- For 2,3-dimethyl-3-pentanol:
- Percent composition = (molar mass of 2,3-dimethyl-3-pentanol / total molar mass) x 100
- Percent composition = (116.23 g/mol / (58.09 g/mol + 116.23 g/mol)) x 100
- Percent composition = (116.23 g/mol / 174.32 g/mol) x 100
- Percent composition = 0.6667 x 100
- Percent composition = 66.67%
Step 3: Calculate the masses of each compound in the sample.
- Multiply the percent composition of each compound by the total mass of the sample.
- For 3-methyl-2-butanone:
- Mass of 3-methyl-2-butanone = percent composition x total mass of the sample
- Mass of 3-methyl-2-butanone = 33.33% x 5.13 g
- Mass of 3-methyl-2-butanone ≈ 1.71 g
- For 2,3-dimethyl-3-pentanol:
- Mass of 2,3-dimethyl-3-pentanol = percent composition x total mass of the sample
- Mass of 2,3-dimethyl-3-pentanol = 66.67% x 5.13 g
- Mass of 2,3-dimethyl-3-pentanol ≈ 3.42 g
Therefore, in the given sample weighing 5.13 g, the mass of 3-methyl-2-butanone is approximately 1.71 g and the mass of 2,3-dimethyl-3-pentanol is approximately 3.42 g.
To determine the masses of 3-methyl-2-butanone and 2,3-dimethyl-3-pentanol in the sample, you need to know their percentage compositions in the sample. Let's denote the mass of 3-methyl-2-butanone as x grams and the mass of 2,3-dimethyl-3-pentanol as y grams.
1. Write down the balanced chemical formulas for both compounds:
- 3-methyl-2-butanone: C5H10O
- 2,3-dimethyl-3-pentanol: C7H16O
2. Calculate the molar masses of both compounds:
- Molar mass of 3-methyl-2-butanone (C5H10O):
Carbon (C) has a molar mass of 12.01 g/mol (5 atoms in the compound)
Hydrogen (H) has a molar mass of 1.01 g/mol (10 atoms in the compound)
Oxygen (O) has a molar mass of 16.00 g/mol (1 atom in the compound)
Total molar mass = (5 x 12.01 g/mol) + (10 x 1.01 g/mol) + (1 x 16.00 g/mol)
= 72.08 g/mol
- Molar mass of 2,3-dimethyl-3-pentanol (C7H16O):
Carbon (C) has a molar mass of 12.01 g/mol (7 atoms in the compound)
Hydrogen (H) has a molar mass of 1.01 g/mol (16 atoms in the compound)
Oxygen (O) has a molar mass of 16.00 g/mol (1 atom in the compound)
Total molar mass = (7 x 12.01 g/mol) + (16 x 1.01 g/mol) + (1 x 16.00 g/mol)
= 130.23 g/mol
3. Calculate the number of moles of each compound in the sample:
- Moles of 3-methyl-2-butanone (x moles) = x (g) / 72.08 (g/mol)
- Moles of 2,3-dimethyl-3-pentanol (y moles) = y (g) / 130.23 (g/mol)
4. Set up a system of equations based on the total mass and moles in the sample:
- x (moles) + y (moles) = total moles
- x (g) + y (g) = total mass
5. Solve the system of equations:
- 72.08x + 130.23y = 5.13 (equation 1)
- x + y = total moles (equation 2)
6. Since the total mass of the sample is given as 5.13 g, we can substitute this value into equation 1 and solve for x and y.
Please provide the value for the total moles to proceed with the calculations.