I have no idea where to begin. J.9 You are asked to identify compound X, which w
ID: 717178 • Letter: I
Question
I have no idea where to begin.
J.9 You are asked to identify compound X, which was extracted from a plant seized by customs inspectors. You run a number of tests and collect the following data. Compound X is a white, crystalline solid. An aqueous solution of X turns litmus red and conducts electricity poorly, even when X is present at appreciable concentrations. When you add sodium hydroxide to the solution a reaction takes place. A solution of the products of the reaction conducts electricity well. An elemental analysis of X shows that the mass percentage composition of the compound is 26.68% C and 2.239% H, with the remainder being oxygen. A mass spectrum of X yields a molar mass of 90.0 g mol"1. (a) Write the empirical formula of X. (b) Write the molecular formula of X (c) Write the balanced chemical equation and the net ionic equation for the reaction of X with sodium hydroxide. (Assume that X has two acidic hydrogen atoms.)Explanation / Answer
a) The percentage composition of X is:
C = 26.68%
H = 2.239%
O = 100 - (26.68+2.239) = 71.081%
Let us assume that we have 100 g of X. Then, the corresponding weights of C, H and O will be:
C = 26.68 g; H = 2.239 g; O = 71.081 g
Now, the atomic mass of C = 12 g/mol; H = 1 g/mol and O = 16 g/mol
Number of moles of each element would be:
# moles of C = 26.68 g/12gmol-1 = 2.223
# moles of H = 2.239/1 = 2.239
# moles of O = 71.081/16 = 4.443
In order to deduce the empirical formula calculate the relative amounts (ratio) of C, H, O
C = 2.223/2.223 = 1
H = 2.239/2.223 = 1.007 = 1
O = 4.443/2.223 = 1.998 = 2
Therefpre, empirical formula = CHO2
b) Empirical formula mass of CHO2 = 12 + 1 + 2(16) = 45 g/mol
Given, molar mass = 90 g/mol
The ratio, 'n' = 90/45 = 2
Molecular formula = n * Empirical formula = 2*(CHO2) = C2H2O4
c) The compound X is oxalic acid (H2C2O4)
Reaction with NaOH
H2C2O4 (aq) + 2NaOH(aq) --------- Na2C2O4(aq) + 2H2O (l)
Net ionic equation:
2H+ (aq) + 2OH-(aq) ----- 2H2O (l)