24.5g of CH3CH2CH2Br was reacted with ammonia to form CH3CH2CH2NH2 at a 75.0% yield, calculate the mass of the product formed.
The first step is to calculate the moles of the reactant CH3CH2CH2Br. To do this you need the formula:moles = m/Mr where m is the mass in grams and Mr is the molar mass which can be worked out using the periodic table.Standard atomic weight of carbon = 12.011Standard atomic weight of hydrogen = 1.008Standard atomic weight of bromine = 79.904There are three carbon atoms, seven hydrogen atoms and one bromine atom. This means that the molar mass is (3 x 12.011) + (7 x 1.008) + (1 x 79.904) = 122.993Because 24.5 g of the reactant was used the moles = 24.5 / 122.993 = 0.20 molesThe next stage is to work out the mass of product with an assumption of a 100% yield. To do this the same formula is used: moles = m/MrThe molar mass of the product needs to be worked out: three carbon atoms, nine hydrogen atoms and one nitrogen atom where the standard atomic weight of nitrogen = 14.007(3 x 12.011) + (9 x 1.008) + (1 x 14.007) = 59.112The formula moles = m/Mr is then rearranged to make the mass the subject, generating the formulam = moles x Mr Using the calculated moles and the molar mass you get: 59.112 x 0.20 = 11.8 gThis is the mass if there was 100% yield, however, this question asked what the mass would be if there was a 75.0% yield, therefore we need to find 75% of this mass: 11.8 x (75/100) = 8.85 g
