# Percentage By Mass Calculation

Submitted by S.Student on Thu, 09/28/2017 - 03:52

A 4.675g sample of limestone (impure CaCO3) was treated with 375ml of 0.261 molL^-1 hydrochloric acid. After the reaction was over, the mixture was filtered and a 25 ml sample of the filtrate required 17.62ml of 0.0517 molL^-1 barium hydroxide solution for complete neutralisation.

Calculate the percentage by mass of CaCO3 in the limestone sample.

Thinking:

I know that the method to solve these types of questions is to work backwards, but since there are technically 2 chemical reactions in the question, I'm confused as to how to relate one to the other in order to solve it.

Characterize the condition for mass percent of a compound. The essential recipe for mass percent of a compound is: mass percent = (mass of synthetic/add up to mass of compound) x 100. You should duplicate by 100 toward the conclusion to express the incentive as a rate.
Duplicate the masses by the mole proportion. Distinguish what number of moles (mole proportion) of every component are in your concoction mixes. The mole proportion is given by the subscript number in the compound. Increase the molecular mass of every component by this mole ratio.
Case 1: Hydrogen has a subscript of two while oxygen has a subscript of 1. Consequently, increase the molecular mass of Hydrogen by 2, 1.00794 X 2 = 2.01588; and leave the molecular mass of Oxygen as seems to be, 15.9994 (duplicated by one).
Case 2: Carbon has a subscript of 6, hydrogen, 12, and oxygen, 6. Duplicating every component by its subscript gives you:
Carbon (12.0107*6) = 72.0642
Hydrogen (1.00794*12) = 12.09528
Oxygen (15.9994*6) = 95.9964