Question

Question:1
You dissolve 312 mg of a compound (A_wB_xC_yD_z) in water producing 14.31 mL of solution and use the photometer and standard solutions to find that the concentration of the A is 1.24 x 10^-2 M. The molar mass of A is 114.52 g/mol.

Moles of A in the solution sample:________________________________

Mass of A in the solution sample: _________________________________

Mass % of A in the compound: _______________________________

Moles of A in 100 grams of the compound: ________________________________



Question:2
The compound, A_wB_xC_y, like all compounds has an oxidation number of 0 (zero). The oxidation number of A is +1, the oxidation number of B is +2, and the oxidation number of C is –1. If there are two moles of A for every 3 moles of C in one mole of the compound, what is the empirical formula of the compound?



Question:3
Given the compound, K_wAl_x(SO₄)_y(H₂O)_z. 100.00 grams of the compound contains 8.24 grams of potassium and 5.69 grams of aluminum. What is the formula of the compound?

Answer 1

weight of compound (AwBxCyDz) = 312 mg

volume in water = 14.31 mL

concentration of the A = 1.24 x 10^-2 M.

means 1.24 x 10^-2 moles in 1000ml water but we have 14.31 ml

moles of A will be there

1) Moles of A in 2.72 Moles of compound (AwBxCyDz) = 17.7 *10^-5 moles

2)

The molar mass of A is 114.52 g/mol.

Mass of A in the solution sample = number of moles * molar mass = 17.7 *10^-5 * 114.52 = 20.27 * 10^-3

3) Weight of compound in the solution sample = 312*10^-3g

Mass % of A in compound = (Mass of A / Total mass ) *100

= g = 6.4%

4) Weight of A in 100 grams of the compound = 6.4g

moles of A in 100 gram of compound = 6.4 / molar mass = 6.4 / 114.52 = 0.055moles

Second question

compound = AwBxCy

two moles of A for every 3 moles of C in one mole of the compound means

But for every 2A there will be 3 C

So for every 1 A There will be 3/2C in that case ABC_3/2

But compound is neutral A is +1 and B is +2 and C is -1

in that case answer will be A₄BC₆

How ever, how many subpart will be there doesnt matter we have been told to answer first 4 question

I am happy to help but i have answered more than that (time is precious for us also)