Unit 5 Chemical Composition
Counting by Mass Individually mass a few Calculate the average mass of one Can count large numbers of by mass
Atomic Mass Unit (amu) 1 amu = 1.66 x 10-24 g
Subatomic particles Particle amu mass (g) Atomic charge Electrical charge (coulomb) Proton (p + ) Electron (e - ) 1.0073 1.673 x 10-24 +1 +1.602 x 10-19.0005486 9.109 x 10-28 -1-1.602 x 10-19 Neutron (n) 1.0087 1.675 x 10-24 0 0
Periodic Table Atomic mass 12.0111 C Element symbol Atomic number 6 Elements with the atomic mass in ( ) means that very little of it exists at any given time. Scientists have not been able to attain an good average atomic mass.
% Composition by Mass When given masses and asked to find the % compositions % Composition of element by mass = grams of element grams of compound ( ) 100
In a 15.0 gram sample of H 2 O, there is 1.68 g H, and 13.32 g O. Find the % composition of each element.
12 grams of carbon monoxide are analyzed and found to contain 5.14 grams of carbon and 6.86 grams of oxygen.
Because atoms and molecules are too small to work with, chemists have defined a large unit called a mole. The mole is defined as the number of atoms in 12 grams of Carbon. 6.02 x 10 23 particles. Avogadro s number = 6.02 x 10 23 (particles: atoms, molecules, ions, electrons, etc.)
Molar mass of an atom, is the atomic mass in grams which is 6.02 x 10 23 atoms. What is the molar mass of iron, of sulfur. Molar mass of a compound is the sum of the atomic mass of all the atoms in the compound in grams which is 6.02 x 10 23 molecules. Calculate the molar mass of hydrochloric acid (HCl) Calculate the molar mass of silver nitrate (AgNO 3 )
Molar mass of water = 18.0 g / mol Therefore, one mole of water is 18 grams of water.
Calculate the number of moles in 12.74 grams of Vanadium. Calculate the number of moles in 12.51 grams of calcium carbonate (CaCO 3 ). moles = grams_ 1 molar mass
Calculate the number of grams in: 0.65 moles of barium iodide, BaI 2 4.0 moles of copper nitrate, Cu(NO 3 ) 2
% composition of element by molar mass 1. Find the molar mass of the compound 2. Divide the molar mass supplied by one element, by the molar mass of the compound. Multiply by 100 to obtain percentage 3. Repeat step 2 for each element % composition of = element by molar mass total molar mass of element ( ) 100 total molar mass of compound
Calculate the % composition of each element in sodium bicarbonate (NaHCO 3 )
Mole to mole ratio in reactions Copper sulfate & iron yields copper & iron sulfate CuSO 4 (aq) + Fe (s) Cu (s) + FeSO 4 (aq) Penny Lab: HCl (aq) + Cl 2(aq) + H 2 (g)
Number of particles = (moles)(6.02 x 10 23 ) Particles = atoms, or molecules, or ions, or electrons, etc.
A hydrate is a molecule with water attached to the crystalline structure. Example: BaCl 2. 2H2 O An anhydrous molecule has NO water attached. % composition water = Molar mass of all water ( ) 100 Molar mass of hydrate Calculate the % water in BaCl 2. 2H2 O
Determine the % composition of H 2 O in the hydrated compound CuSO 4. 5H 2 O.
Hydrate Lab Determine the % composition of H 2 O for one of the samples. (Teacher will assign the sample) Bonus point for determining the number of H 2 O molecules attached to each molecule.? = moles of water moles of anhydrous
Molecular formula is the actual formula that gives the exact composition. Example: C 6 H 12 O 6 (glucose sugar) Empirical formula is the smallest whole number ratio of the atoms present. Example: CH 2 O (empirical formula for glucose) If multiplied by 6, it is the formula for glucose.
Steps for establishing formula 1. List mass of elements (if %, use mass in 100 grams) 2. Convert mass of element to moles 3. Write a tentative formula using mole values as subscripts 4. Attempt to convert subscripts to smallest whole numbers by dividing mole values by the smallest mole value 5. If decimal numbers, multiply by a small whole number to obtain empirical formula 6. To find molecular formula, compare empirical formula mass to molar mass of compound. Multiply subscripts by multiplication factor.
A compound contains 4.151 g if aluminum and 3.692 g of oxygen. Determine the empirical formula
A compound having an approximate molar mass of 168 g/mol has the following percentage composition by mass; carbon = 42.87%, hydrogen = 3.598%, oxygen = 28.55 %, and nitrogen = 25.00 %. Determine the empirical formula. Determine the molecular formula.
Review 1. a. 32.046 g / mol b. 158.032 g / mol c. 310.18 g / mol 2. a. 0.158 mole SO 2 b. 0.148 mole NaBr c. 0.0777 mole Ba(NO 3 ) 2 3. a. 327.52 g HNO 3 b. 79.22 g CO 2 c. 363.82 g Ca(C 2 H 3 O 2 ) 2 4. a. 9.03 x 10 23 molecules b. 1.75 x 10 24 molecules c. 2.71 x 10 22 molecules d. 7.76 x 10 21 molecules 5. a. 43.38 % Na 45.29% O 11.33 % C b. 21.21 % N 24.27 % S 6.10 % H 48.44 % O 6. a. 63.52 % Ag 8.16 % N 28.32 % O 7. 51.16 % H 2 O 8. (Empirical formula problem)
Mole to mole ratio in reactions Copper sulfate & zinc yields copper & zinc sulfate CuSO 4 (aq) + Zn (s) Cu (s) + ZnSO 4 (aq)