Question

What is the mass of a `1.26xx10^-4*mol` quantity of acetic acid?

Answer 1

`"In acetic acid,"` `H_3"CCO"_2H`?

Explanation:

Now, the basic equation is `"moles"` `=` `"mass"/"molar mass"`....and thus `"mass"` `=` `"moles"` `xx` `"molar mass"`.

So here,

`"mass"` `=` `1.26xx10^-4*cancel(mol)xx60.05*g*cancel(mol^-1)=7.60*mg`.

Certainly, this expression is dimensionally consistent. I wanted an answer with units of mass, and the calculation did in fact give such units.

Answer 2

The mass of `"CH"_3"COOH"` is `"0.00757 g"`.

Explanation:

Multiply moles by molar mass.

The molar mass of acetic acid `("CH"_3"COOH")` is `"60.052 g/mol"`.
https://www.ncbi.nlm.nih.gov/pccompound?term=%22acetic+acid%22

`1.26xx10^(-4)color(red)cancel(color(black)("mol CH"_3"COOH"))xx(60.052"g CH"_3"COOH")/(1color(red)cancel(color(black)("mol CH"_3"COOH")))="0.00757 g CH"_3"COOH"` (rounded to three significant figures)