September

 Molar mass – the mass of one mole of a substance ◦ Gram formula mass (gfm) – mass in grams of one mole of an ionic compound ◦ Gram molecular mass (gmm) – mass in grams of one mole of a molecular compound  To calculate molar mass… ◦ Add the atomic masses of each atom in a compound and switch the label from AMU to gram

 Find the gram formula mass of beryllium oxide ◦ Beryllium oxide = BeO Be = 9.0 AMU O = 16.0 AMU Total = 25.0 AMU -> 25.0 grams ◦ Lithium sulfide = Li 2 S ◦ Li = 4.0 AMU x 2 = 8.0 AMU ◦ S = 32.1 AMU  = 40.1 AMU -> 40.1 g

 If you have 100 coins and 30 of them are quarters, what percent of all your coins is quarters? percent = part/whole x 100 part = 30 and whole = /100 x 100= 30%

Percent Composition: the percent by mass of each element in a compound To calculate percent composition, find the mass of the element and divide that by the total mass of the compound. % composition = mass of element x100 mass of compound In a sample of hydrogen peroxide, the mass of hydrogen is 2.0 grams and the total mass of the compound is 34.0 grams. Find the percent composition of hydrogen in hydrogen peroxide.

 8.20 g of magnesium combines with 5.40 g of oxygen to form a compound.  Find the percent composition of this compound…  Total mass = 8.20g g = 13.6g  %Mg = 8.20g/13.6g x100 = 60.3%  %O = 5.40g/13.6g x100 = 39.7% ◦ Check… 60.3% % = 100% 

 29.0 g Ag combine with 4.30 g S to form a compound. Find the percent composition. ◦ Hint… don’t forget to find the total mass of the whole compound! ◦ Total mass = 29.0g g = 33.30g ◦ %Ag = 29.0g/33.30g x100 = 87.1% ◦ %S = 4.30g/33.30g x100 = 12.9%  Check…. 87.1% % = 100% 

 Propane is C 3 H 8. Use molar mass to find percent composition. ◦ Use the periodic table to find the atomic mass of each element… ◦ C=12.0AMU and H = 1.0AMU and change that to molar mass ◦ C = 12.0g and H = 1.0g ◦ Three carbon atoms in the formula = 12.0g x 3 = 36.0g ◦ Eight hydrogen atoms = 1.0g x 8 = 8.0g ◦ Total molar mass of compound = = 44.0g ◦ %C = 36.0g/44.0g x100 = 81.8% ◦ %H = 8.0g/44.0g x100 = 18.2% ◦ Check… 81.8% % = 100% 

 What is the mass of carbon in 82.0 g propane? ◦ Hint… use carbon’s percent composition! ◦ We just found that the percent of carbon in propane is 81.8%. That means that 81.8% of the mass of a sample of propane will be carbon. ◦ So to calculate the mass of carbon in 82.0 g, just multiply by the percent, 81.8%. Don’t forget to change the percent to a decimal! ◦ 82.0g x = 67.1 g carbon  That means that in 82.0g propane, there are 67.1 g carbon

 Empirical Formula: formula with the lowest whole- number ratio of elements in a compound ◦ In other words… reduce the fraction! ◦ Glucose : C 6 H 12 O 6 ratio is 6:12:6… you can divide each of those by 6 and get 1:2:1 so the empirical formula is C 1 H 2 O 1 or CH 2 O ◦ Benzene: C 6 H 6 ratio is 6:6, divide both by 6 and get 1:1…. CH ◦ Ribose: C 5 H 10 O 5 ratio is 5:10:5, divide each by 5 and get 1:2:1…. CH 2 O

S 2 Cl 2 divide both by 2… SCl C 6 H 10 O 4 divide all by 2, C 3 H 5 O 2 H 2 O 2 HO

 Find percent composition ◦ Carbon is black, oxygen is red, hydrogen is white ◦ Use the periodic table to mind atomic mass…. C=12.0AMU, O=16.0AMU, H=1.0AMU ◦ Count the atoms to write the chemical formula… C 3 H 6 O ◦ Molar mass: C=12.0g x 3 = 36.0g ◦ H=1.0g x 6 = 6.0g ◦ O = 16.0g ◦ Add all the masses to find total gram molecular mass… 36.0g+6.0g+16.0g=58.0g ◦ %C=36.0g/58.0g = 62.1% ◦ %H=6.0g/58.0g = 10.3% ◦ %O=16.0g/58.0g = 27.6% ◦ Check… 62.1%+10.3%+27.6%=100%   Write the empirical formula of acetone… C 3 H 6 O… can’t be reduced