Using Chemical Formulas

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Presentation transcript:

Using Chemical Formulas

Definitions The Mole Avogadro’s Number The amount of a substance that contains as many particles as there are atoms in exactly 12g of carbon-12. Avogadro’s Number The number of particles in exactly one mole of a pure substance. 6.022 x 1023

Understanding the Formula Relationship between the number of moles of each atom in a chemical compound H2SO4 2 moles of Hydrogen 1 mole of Sulfur 4 moles of Oxygen

Molar Mass The mass of one mole of a pure substance. 1 mole Water = 18.00g 1 mole ammonia = 17.00g 1 mole Carbon Dioxide = 44.00g

Molar Mass Molar Mass Calculation for Sulfuric Acid (H2SO4) Make a list of each element and the number of atoms of each element present in the substance. H 2 S 1 O 4

Molar Mass Go to periodic table and determine the atomic mass average (atomic weight) of each element. H 1.00794 g/mol S 32.066 g/mol O 15.9994 g/mol

Molar Mass Multiply each atomic mass by the number of atoms in the formula. H 1.00794 * 2 = 2.015 g/mol S 32.066 * 1 = 32.066 g/mol O 15.9994 * 4 = 63.998 g/mol

Molar Mass Add up the results of step three: 2.015+32.066+63.998 = 98.079 g/mol = molar mass of sulfuric acid

Molar Mass Calculations The Four Steps to Solving Mole Problems: Step 1: Figure out how many parts your calculation will have by using the diagram

Molar Mass Calculations Step 2: Make a T-chart, and put whatever information the problem gave you in the top left. After that, put the units of whatever you were given in the bottom right of the T, and the units of what you want to find in the top right.

Molar Mass Calculations Step 3: Put the conversion factors into the T-chart in front of the units on the right. As we saw from the "map", the conversion factor between grams and moles is the atomic mass of copper. Because we measure atomic mass in grams, you need to put the atomic mass in front of the unit "grams of copper". What do you put in front of moles? Whenever you do a calculation of this kind, you need to put "1" in front of moles, like you see here:

Molar Mass Calculations Step 4: Cancel out the units from the top left and bottom right, then find the answer by multiplying all the stuff on the top together and dividing it by the stuff on the bottom. In this case, you'd multiply 22 by one and divide the result by 63.5. Your answer, 0.35 moles of copper:

Molar Mass Calculations Solving Two-Step Mole Calculation Problems: What happens if we need to solve a problem that requires we not just go from one box in the next in our diagram, but across the entire diagram?

Molar Mass Calculations How many atoms are in 22 grams of copper? Instead of doing a simple one step calculation, we need to do a two-step calculation, with the first step going from grams to moles and the second step going from moles to atoms.

Molar Mass Calculations How can we solve this kind of problem? We start off by doing the same thing that we did in our last example: We had to convert grams to moles before, and we can see from the map that we have to convert grams to moles now, too. To refresh your memory, here's the calculation from last time:

Molar Mass Calculations In the next step, we do the same thing over again, except that we need to add another T to the T-chart. When you do this, take the units of the thing at the new top left and put them on the bottom right (in this case, moles). Then take the units of what you want (in this case, atoms) and put it in the top right. Finally, put in your conversion factors, which from the chart above is Avogadro's number, or 6.022 x 1023. Since this number refers to the number of atoms in a mole of a substance, we put this in front of "atoms of copper". Again, put the number "1" in front of moles, because we're saying that there are 6.022 x 1023 atoms in ONE mole of an element.

Molar Mass Calculations When we add all these terms in, we can cross out the units that cancel out, as shown. To get the answer, multiply all the numbers on the top together and divide by the numbers on the bottom. Your answer should then be set up like this: