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.
. 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?
1 answer