Aim: How to determine Empirical and Molecular Formula

•DO NOW:Here is data from an experiment: 1. Mass of empty crucible + cover = 11.70 g2. Mass of crucible + cover + hydrated crystal (before

heating) = 14.90 g3. Mass of crucible + cover + anhydrous salt (after heating)

= 14.53 gWhat is the % by mass of the water in the hydrated crystal?

How to determine empirical and molecular formula

Empirical Formula vs. Molecular Formula

•Empirical Formula tells us what elements are present in the compound in the simplest whole number ratio of elements.•Molecular formula tells which elements are present in the compound and the actual number of each.

How to determine empirical and molecular formula

Empirical Formula from Masses of Elements

•A sample weighing 1.587 g of a compound contains 0.483 g N and 1.104 g O. What is the empirical formula of the compound?•This compound has the formula NxOy.•We need to convert the mass of N and O to moles of N and O. •Then divide each by the smallest mole to get a whole number ratio.

Solution

0.483 g N x = 0.0345 mol N

1.104 g O x = 0.06900 mol O

How to determine empirical and molecular formula

Solution

N = = 1.00

O = = 2.00

You divide each mole by the smallest one (0.0345). For N you get 1.00 and for O you get 2.00. The empirical formula is NO2

Empirical Formula from Percent Composition

•An analysis of sodium dichromate gives the following percentages: 17.5% Na, 39.7% Cr, and 42.8% O. What is the empirical formula of this compound?•This compound has the formula NaxCryOz. •Assume that you have an 100.0 g sample of the compound. Then the mass of each element in the sample equals the numerical value of the percentage.

Solution

•Of the 100.0g of sodium dichromate, 17.5g is Na, 39.7g is Cr, and 42.8 g is O. You convert these amounts to moles.

17.5 g Na x = 0.761 mol Na39.7 g Cr x = 0.763 mol Cr42.8 g O x = 2.68 mol 0

How to determine empirical and molecular formula

Solution

•Now you divide all mole numbers by the smallest one.

For Na: (0.761 mol)/(0.761 mol) = 1.00For Cr: (0.763 mol)/(0.761 mol) = 1.00For O: (2.68 mol)/(0.761 mol) = 3.52

The subscripts are not all whole numbers. To be made into whole numbers multiply each one by 2 to get an empirical formula of Na2Cr2O7.

How to determine empirical and molecular formula

Molecular Formula from Empirical Formula

•The molecular formula of a compound is a multiple of its empirical formula.•The molecular formula may be determined by dividing the molar mass of the compound by the empirical molar mass. When you divide you find which multiple it is of the empirical formula.

Determining Molecular Formula

•A compound is 80.0 % C and 20.0 % H by mass. If its molecular mass is 75.0 g, what is its empirical formula? What is its molecular formula?

First, we must determine the empirical formula.Step 1: Assume a 100 g sample. Convert grams to moles 80.0 % C = 80.0 g C x = 6.67 mol C 20.0 % H = 20.0 g H x = 20. 0 mol

How to determine empirical and molecular formula

Step 2: Divide each mole number by the smallest mole number and round to the nearest integer

For C: = 6.67/6.67 = 1

For H: = 20.0 / 6.67 = 3

The empirical formula is CH3

How to determine empirical and molecular formula

Step 3: Determine molar mass of empirical formula

CH3 = C + 3H = (12.0 g/mol) + (3 x 1.0 g/mol) = 15.0 g/mol

Step 4: Divide molecular molar mass of compound (75. 0 g/mol) by empirical molar mass.

= = 5

Multiply the empirical formula CH3 by 5. The molecular formula is C5H15.

How to determine empirical and molecular formula

How to determine empirical and molecular formula

Practice

•What is the molecular formula of a compound that has an empirical formula of NO2 and molecular mass of 92.0 g?