The molecular weight of acetic acid is 60.052 g/mol. And 0.5 M NaOH is 0.5 moles/1000 mL NaOH. Set up the stoichiometry as:
3 g HC2H3O2 x 1 mol HC2H3O2 x 1 mol NaOH x 1000 mL NaOH = 99.9 mL NaOH
1 60.052 g HC2H3O2 1 mol HC2H3O2 0.5 mol NaOH
Sarah H.
tutor
So in this problem you have two pieces of information to work with: 3.0 grams of acetic acid and 0.5 M of NaOH. Molarity is a conversion factor so I wait to use this until later on in the problem. I start with 3 grams of acetic acid. To get to moles of acetic acid, you have to divide by the molecular weight. So start with 3 grams of acetic acid divided by 60.052 grams (molecular weight). Then to get to moles of NaOH, you know that acetic acid can only donate one hydrogen per molecule since it is a monoprotic acid and NaOH can only donate one hydroxide so you set up a proportion of the H to OH which can be seen above as 1 mole NaOH divided by 1 mole HC2H3O2 (some people will leave this calculation out unless it's a diprotic/triprotic acid). Then to get to mL, you need to use molarity. We know molarity units are moles/L which can also be expressed as moles/1000 mL (1 L is equal to 1000 mL). Multiply all proportions to find your answer.
Does that help? Is there a certain part that does not make sense?
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05/16/22
Jessica L.
im confused. isnt molecular weight the addition of every elements mass? how is it 60? H = 1*4 , C = 2*12, O = 16*2?
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11/20/22
Janet J.
I don’t understand this05/16/22