A certain vinegar is 6.02% acetic acid (HC2H3O2) by mass. How many grams of HC2H3O2 are contained in a 355-mL bottle of vinegar? Assume a density of 1.01g/mL.
355mL (1.01g/mL) (0.0602) = 21.6g
A certain vinegar is 6.02% acetic acid (HC2H3O2) by mass. How many grams of HC2H3O2 are contained in a 355-mL bottle of vinegar? Assume a density of 1.01g/mL.
To find the number of grams of HC2H3O2 in the 355 mL bottle of vinegar, you need to use the molarity (percentage by mass) and the density of the vinegar.
Step 1: Calculate the volume of acetic acid in the vinegar.
Given:
Volume of vinegar = 355 mL
Density of vinegar = 1.01 g/mL
Volume of acetic acid = Volume of vinegar x Molarity of acetic acid
= 355 mL x (6.02/100)
= 355 mL x 0.0602
= 21.451 mL
Step 2: Convert the volume of acetic acid to grams using the density.
Mass of acetic acid = Volume of acetic acid x Density of vinegar
= 21.451 mL x 1.01 g/mL
= 21.70151 g
Therefore, there are approximately 21.70 grams of HC2H3O2 in a 355 mL bottle of vinegar.
density = 1.01 g/mL so what is the mass of 355 mL?
1.01 g/mL x 355 mL = ?? grams.
How much of that is HC2H3O2? It is 6.02%; therefore,
?? grams x 0.062 = xx grams HC2H3O2.