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Learning Stoichiometry

Victoria Stackpole

on 21 December 2010

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Transcript of Stoichiometry

Learn it! Molar Mass This number is the
atomic weight The to Finding Molar Mass (steps) The first thing that is done to find the molar mass is to look at how many moles each element has. For example, NaO has one mole of Sodium and 3 moles of Oxygen.
Next is multiplication. Let's keep going with the NaO . If Na has one mole, and the molar mass is 22.990 g then we know that 1 times 22.990 is 22.990. Scientifically to do that problem we would start with (moles of Na X molar mass of Na) (1 X 22.990) Now lets focus on the 3 moles of Oxygen. All we have to do is multiply 3 times 15.999. Scientifically, (moles of O X molar mass of O) (3 X 15.999). We know, by doing some basic math that the molar mass of 3 moles of oxygen is 47.997 g of Oxygen. Addition. We have the molar mass of Sodium and the molar mass of Oxygen (3 moles) now all that is left is to add them together. 22.990 + 47.997= 70.987. Scientifically, we would write that as (moles of Na X molar mass of Na)+(moles of O X molar mass of 0)=Molar Mass of NaO 3 3

3 That's all. Easy as pie. What if there is a problem like
Mg (PO ) ? 3 4 2 1. First we know that Mg (Magnesium) has 3 moles. 2. Second we know that O (Oxygen) already has 4 moles.
3. Third, we know that when something is in parenthesis that whatever number is outside of parenthesis to the right, you multiply everything inside by that number. 4. Because that's all that is new, we now have 2 moles of Phosphorus and 8 moles of Oxygen. 5. Now the problem is worked like the previous one we learned. 6. Easy Peasy, Lemon Squeezy Check out the Periodic table for the atomic weight. Check out the Periodic table for the atomic weight. Don't be afraid. (: 1.




5. Get your own piece of paper and figure out the molar mass. H O H PO P O Al(NO ) (NH ) CO 2 3 4 2 5 3 3

2 3 Answers 1.




5. (2 X 1.0079)+(1X15.999)=18.014 molar mass of H o Learn it! Emperical Formula 2 (3 X 1.0079 g/mol H)+(1X30.973 g/mol P)+(4 X 15.999 g/mol O) = 97.993 g/mol H PO 3 4 Empirical Formula-
formula for a molecule that is expressed as the smallest possible whole number ratio of the elements in the compound. So how do we find an empirical formula? 1: Change each element from gram to moles by using the molar mass values from the periodic table.
2. When you get the mole values, divide each by the lowest mole value that you got.
3. After this step, most of your values should be whole numbers. If not you need to multiply to make them whole numbers.
If the value ends in .5, multiply it by 2 to get your whole number.
If it ends in .33, multiply it by 3 to get a whole number. Here's an example: We have 29.0 g of sodium,40.5 g of sulfur, and 30.4 g of oxygen. 29.0/22.99= 1.26 moles Sodium
40.5/32.07= 1.26 moles Sulfur
30.4/16.00= 1.9 moles Oxygen So we divide by the smallest number... 1.26 moles Na/1.26= 1 Na
1.26 moles S/1.26= 1 S
1.9 moles O/1.26= 1.5 O 1 Na *2= 2 Na
1 S *2= 2 S
1.5 O *2= 3O Lets put this all together. Na2S2O3 But what if you get
40.0 % C, 6.67% H, and 53.33% O? Don't sweat it. Just assume it's out of 100%.
Just changed the percents to numbers and follow the steps above. 40.0% C= 40.0 g of C
6.67% H= 6.67 g of H
53.33% O= 53.33 g of O. Try these: 1) 13.5 g Ca, 10.8 g O, and 0.675 g H.
2) 57.14% C, 6.16% H, 9.52% N, 27.18% O.
3) 54.09% Ca, 43.18% O, 2.73% H.
4) 8.65 g Fe, 3.72 g O
5) 207 g Pb, 32 g S Answers:
1) CaO2H2 or Ca(OH)2
2) C14H18N2O5
3) CaO2H2 or Ca(OH)2
4) Fe2O3
5) PbS (2 X 30.975 g/mol P)+(5 X 15.999 g/mol o)=141.94 g/mol P O 2 5 (1 X 26.981 g/mol al)+(3 x 14.006 g/mol n)+(9 + 15.999 g/mol o)=212.99 g/mol al(no ) 3 3 (2 × 14.006 g/mol N)+(8 × 1.0079 g/mol H)+(1 × 12.010 g/mol C)+(3 × 15.999 g/mol O)=96.082 g/mol (NH ) CO 4 2
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