Empirical Formula of a Copper Oxide Compound

A Lab….

Essential Definitions

The Chemical Equation

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• Written expression of a chemical or physical change. Uses chemical formulas to describe individual substances.

Anatomy of the Chemical Equation

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• Chemical equations contain ____________ and ___________.

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Al₂O₃(s) → Al(s) + O(g)

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reactants

products

Anatomy of the Chemical Equation

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Al₂O₃(s) → Al(s) + O(g)

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reactants

products

The substance(s) you start with before the reaction takes place.

The substance(s) you end with after the reaction takes place.

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Empirical Formula

We have seen many empirical formulas in the form on ionic compounds:

Al₂O₃ Li₃PO₄ MgO Be₂O₂

The last compound listed is NOT an empirical formula. Why?

Because it’s not reduced!

Empirical Formula

Empirical formulas are the reduced mole ratio of atoms within a compound.

Law of Definite Proportions!

Substances combine in unique, unchanging, whole number ratios.

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• That means water, H₂O, will always have the same mole ratio: 2 moles of H to 1 mole of O for every molecule.

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• It doesn’t matter if you have 4 grams of water or 40 grams of water.

Law of Definite Proportions!

• You have 15 g of the copper oxide compound and your friend has 20 g of it:
• You both have different starting masses, right?
• BUT you also have the same compound...soooo….

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• The mole ratio will be the same, because you are using the same substance.
• The mass ratio will ALSO be the same because you are using the same substance.

The Empirical Formula Lab

Objective

In our lab, we are trying to figure out the empirical formula of a substance that contains copper and oxygen.

What are we doing?

• We start with some copper oxide of an unknown formula.
• It could be CuO OR Cu₂O because copper is a transition metal and has more than one possible charge.

What are we doing?

• In order to figure out if we are working with CuO or Cu₂O, we need to remove the oxygen and then calculate the mole ratio.

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What is a mole ratio?

• The MOLE RATIO is how many moles of each element are present in the compound.

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• Ex: CuO has a 1:1 mole ratio (1 mol of Cu for every 1 mol of O)

Cu₂O has a 2:1 mole ratio (2 mol of Cu for every 1 mol of O)

How are we doing that?

• We are releasing the oxygen from the copper oxide and test tube by filling the test tube with gas and igniting it with the oxygen.
• This removes the oxygen from the compound and the test tube.

How are we doing that?

Knowing that we are removing oxygen from the compound, think about what the anticipated mass of the product would be.

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Heavier? Lighter?

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*Consider how lab errors may impact these results.

Calculations in the Empirical Formula Lab

We are using an EXAMPLE.

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DO NOT COPY THE NUMBERS INTO YOUR LAB. THEY ARE A MODEL!

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THE CALCULATIONS ARE WHAT YOU ARE USING AS A GUIDE!

Here’s our data for Al_xO_y(s)→Al(s)+O₂(g)

Calculate the mass of the Al_xO_Y sample

Subtract the mass of the test tube from the mass of the test tube and starting sample.

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42.093 g - 34.089 g =

8.004 g Al_xO_y

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8.004 g

Calculate the mass of Al in the Al_xO_Y sample

Subtract the mass of the test tube from the mass of the test tube and ending sample.

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38.326 g - 34.089 g =

4.237 g Al

in Al_xO_y

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FYI: We got this number when we took the mass of the test tube when we finished heating the sample.

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4.237 g

Calculate the mass of O in the Al_xO_Y sample

Subtract the mass of Al from the mass of the Al_xO_y sample.

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8.004 g Al_xO_y - 4.237 g Al =

3.767 g O

in Al_xO_y

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3.767 g

Calculate the MOLES of Aluminum (Al)

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Start: grams of aluminum.

Unknown: moles of aluminum.

Conversion(s): 1 mol of X = MM of X (g)

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1 mol Al

26.98 g Al

= 0.1570422535 → 0.1570 mol Al

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4.237 g Al x

* Molar Mass

From the p-table

Calculate the MOLES of Oxygen (O)

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Start: grams of oxygen.

Unknown: moles of oxygen.

Conversion(s): 1 mol of X = MM of X (g)

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1 mol O

16.00 g O

= 0.2354375 mol O → 0.2354 mol O

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Needed 4 Sig Figs

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3.767 g O x

*Molar Mass

From the p-table

Find the MOLE RATIO to determine the final formula for compound

REMINDER: The same number of Al always bonds with the same name of O when forming this compound! (Law of Definite Proportions)

0.2354 mol O

0.1579 mol Al

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Mole Ratio =

= 1.5 mol of O for every 1 mol Al

(should be a whole number or really, really close to a whole number...! SIG FIGS DON’T MATTER HERE.)

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Refer to the EF / MF Cheat Sheet given to you in Unit 05. This will remind you how to calculate the MOLE RATIO & EMPIRICAL FORMULAS

Find the MOLE RATIO to determine the final formula for compound

1.5 mol of O for every 1 mol Al

x 2

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Most of the time you will end up with a whole number. If you do, use this as your ratio. If you have a # right in the middle, like mine, multiply the ratio by 2.

= 3 mol of O for every 2 mol Al

Let’s complete our formula!

Al₂O₃

Use the MOLE RATIO!

3 mol of O for every 2 mol of Al

Percent Composition: Refresher

Percent Composition

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Molar mass of the element

Molar mass of the compound

= % composition of element

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x 100

Experimental mass of the element

Experimental mass of the compound

= % composition of element

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x 100

*REMEMBER! There are two ways to calculate percent composition