1) divide each percentage by the relative atomic mass of the element
2) divide all results by the smallest number
3)multiply by a whole number to get the simplest whole number ratio (if necessary)
that is to say:
Na S O
32.37÷23 22.58÷32 45.05÷16
= 1.407 = 0.7056 = 2.816 (to 4 significant figures)
the smallest number here is 0.7056 so:
1.407÷0.7056 0.7056÷0.7056 2.816÷0.7056
=1.99 approx.2 = 1 3.99 approx. 4
here there is no need to carry out step 3 as ratio obtained is already a simplest whole number ratio
so empirical formula is: Na₂SO₄
Answer:
32.6%
Explanation:
Equation of reaction
2KClO₃ (s) → 2KCl (s) + 3O₂ (g)
Molar mass of 2KClO₃ = 245.2 g/mol ( 122.6 × 2)
Molar volume of Oxygen at s.t.p = 22.4L / mol
since the gas was collected over water,
total pressure = pressure of water vapor + pressure of oxygen gas
0.976 = 0.04184211 atm + pressure of oxygen gas at 30°C
pressure of oxygen = 0.976 - 0.04184211 = 0.9341579 atm = P1
P2 = 1 atm, V1 = 789ml, V2 = unknown, T1 = 303K, T2 = 273k at s.t.p
Using ideal gas equation
=
V2 =
V2 = 664.1052 ml
245.2 yielded 67.2 molar volume of oxygen
0.66411 will yield =
= 2.4232 g
percentage of potassium chlorate in the original mixture =
= 32.6%