In: Chemistry
Question:1
You dissolve 312 mg of a compound
(AwBxCyDz) 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, AwBxCy, 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,
KwAlx(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?
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 ABC3/2
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)