A compound is found to contain 31.42 % sulfur, 31.35 % oxygen, and 37.23 % fluorine by weight. To answer the questions, enter the elements in the order presented above. 1. What is the empirical formula for this compound? 2. The molecular weight for this compound is 102.1 g/mol. What is the molecular formula for this compound?

Explanation:

To obtain the empirical and molecular formula of this compound from the percent composition of the elements, we follow the steps below;

Step 1: Divide the percentage composition by the atomic mass

Sulphur = 31.42 / 32 = 0.9819

Oxygen = 31.35 / 16 = 1.9594

Flourine = 37.23 / 19 = 1.9595

Step 2: Divide by the lowest number

Sulphur = 0.9819 / 0.9819 = 1

Oxygen = 1.9594 / 0.9819 ≈ 2

Flourine = 1.9595 / 0.9819 ≈ 2

This means the ratio of the elements is 1 : 2: 2

The empirical formular (simplest formular of a compound) of the compound is;

SO₂F₂

To obtain the molecular formular (Actual formular of a compound);

(SO₂F₂)n = 102.1

Inserting the atomic masses and solving for n;

(102)n = 102.1

n ≈ 1

The molecular formular is; (SO₂F₂)₁ = SO₂F₂

a is the answer to this question

explanation:

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :