CH3COOH + C5H12O -> C7H14O2 + H2O
If the yield for the reaction is 45%, how many grams of the product are formed when 3.58 g of acetic acid are reacted with 4.75 grams of isopentyl alcohol?
Millimoles of acetic acid= 59.6667
Millimoles of isopentyl alcohol= 53.9773
How would I convert mmoles acetic acid to mmoles of the product and mmoles isopentyl alcohol to mmoles of the product?
Would I take the millimoles of each(acetic acid and isopentyl alcohol) and multiply that value by the molecular weight of the products (148 g/mol)?
To convert millimoles of acetic acid and isopentyl alcohol to millimoles of the product, you need to use the stoichiometry of the balanced chemical equation:
CH3COOH + C5H12O -> C7H14O2 + H2O
From the equation, you can see that 1 mole of acetic acid (CH3COOH) reacts with 1 mole of isopentyl alcohol (C5H12O) to produce 1 mole of the product (C7H14O2).
So, to convert millimoles of acetic acid (CH3COOH) to millimoles of the product (C7H14O2), you multiply the millimoles of acetic acid by the stoichiometric ratio:
Millimoles of product = Millimoles of acetic acid × (1 mole product)/(1 mole acetic acid)
Similarly, to convert millimoles of isopentyl alcohol (C5H12O) to millimoles of the product (C7H14O2), you multiply the millimoles of isopentyl alcohol by the stoichiometric ratio:
Millimoles of product = Millimoles of isopentyl alcohol × (1 mole product)/(1 mole isopentyl alcohol)
Once you have the millimoles of the product, you can then convert it to grams by multiplying by the molar mass of the product. In this case, you would multiply the millimoles of the product by the molar mass of C7H14O2 (148 g/mol) to get the grams of the product.