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 Based on lab evidence  For example, we know that 4.9 g of magnesium react with 32.0 g of Bromine.  How can we figure out the formula of the compound.

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Presentation on theme: " Based on lab evidence  For example, we know that 4.9 g of magnesium react with 32.0 g of Bromine.  How can we figure out the formula of the compound."— Presentation transcript:

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2  Based on lab evidence  For example, we know that 4.9 g of magnesium react with 32.0 g of Bromine.  How can we figure out the formula of the compound that is formed?

3 1. Convert the mass to moles for each element. If percents are given, just change the % to grams. 4.9 g Mg =.20 mol Mg 32.0 g Br =.400 mol Br 1 mol Mg g Mg X _______________ 1 mol Br g Br

4 2. Divide all of the moles by the lowest number of moles. 4.9 g Mg =.20 mol Mg/.20 mol = g Br =.400 mol Br/.20 mol = 2 1 mol Mg g Mg X _______________ 1 mol Br g Br

5 3. a. If all of your numbers are whole numbers (or really close ~ ), that is how many atoms there are in the formula. 4.9 g Mg =.20 mol Mg/.20 mol = g Br =.400 mol Br/.20 mol = 2 So you have 1 Mg and 2 Br, so it’s MgBr 2 1 mol Mg g Mg X _______________ 1 mol Br g Br

6 3. b. If any of your answers are ___.5 (or really close ~ ), then multiply ALL the numbers by 2 so they’re all whole numbers. That is how many atoms are in the formula.

7  Example problem: Find the empirical formula of a compound that is 70.0% iron and 30.0% oxygen. Step 1: 70.0 g Fe 30.0 g O X _______________ 1 mol Fe g Fe = 1.25 mol Fe X _______________ 1 mol O g O = 1.88 mol O

8 Steps 2 & 3: 70.0 g Fe 30.0 g O X ___________ 1 mol Fe g Fe = 1.25 mol Fe/1.25 mol = 1x2=2 X ___________ 1 mol O g O = 1.88 mol O/1.25 mol = 1.5x2 =3 Fe 2 O 3

9 1. Find the empirical formula. 2. Find the molar mass of the empirical formula. 3. The problem will tell you the molar mass of the molecule. Take this given mass and divide it by the molar mass of the emp. form. ***This should always be a whole number. 4. Multiply the subscripts in the empirical formula by this number to get the molecular formula.


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