A 1.000g mixture of As4S6(s) and As4S4(s) was burned in excess O2. The products were As4O6(s) and SO2(g). The solid product had a mass of 0.905g. Find the mass % of As4S4(s) in the original mixture.

Answer 1

These problems are two equations in two unknowns that are solved simultaneously.
Let X = mass As4S4
and Y = mass As4S6
------------------------
equation 1 is X + Y = 1.000

The second equation comes from the 0.905 g of the As4O6 product formed.
The equations are
As4S4 + 7O2 ==> As4O6 + 4SO2
As4S6 + 9O2 ==> As4O6 + 10SO2
To save on typing space I'll lt mm stand for molar mass and as stand for atomic mass.

As4O6 formed from As4S4 is
X(mm As4O6/mm As4S4)

As4O6 formed from As4S6 is
Y(mm As4O6/mm As4S6)

If we add those together we come up with equation 2 which is
X(mm As4O6/mm As4S4) + Y(mm As4O6/mm As4S6) = 0.905 g since all of the As4O6 formed has a mass of 0.905 g.

Solve the two equations, find X (you don't need to do Y unless you're interested), then
%As4S4 = (mass As4S4/mass sample)*100 = (X/1.000)*100 = ?

Post your work if you get stuck.

Answer 2

83% as4s4