# Question ffd21

May 22, 2015

Your unknown acid has a molar mass of 90.13 g/mol.

You don't even need to write a balanced chemical equation for this neutralization reaction, since all you really need to know is the mole ratio that exists between the acid and sodium hydroxide.

The $1 : 2$ mole ratio given tells you that you can use the number of moles of sodium hydroxide that were needed to neutralize the acid to determine how many moles of the acid were present in the original solution.

Use the molarity and volume of the sodium hydroxide solution to determine how many moles you have

$C = \frac{n}{V} \implies n = C \cdot V$

${n}_{N a O H} = \text{0.300 M" * 12.5 * 10^(-3)"L" = "0.00375 moles}$ $N a O H$

Now use the mole ratio to figure out the number of moles of ${H}_{2} X$ that reacted

0.00375cancel("moles"NaOH) * ("1 mole"H_2X)/(2cancel("moles"NaOH)) = "0.001875 moles" ${H}_{2} X$

Since you the mass of the sample, you can determine the acid's molar mass by

M_M = m/n = "0.169 g"/"0.001875 moles" = color(green)("90.1 g/mol")#

The closest match I could find is oxalic acid, ${H}_{2} {C}_{2} {O}_{4}$, which has a molar mass of $\text{90.03 g/mol}$.