Unit 3: Quantities and Chemical Reactions

Unit 3: Quantities and Chemical Reactions Precision - How certain you are of a measurement - How consistent the measuring device is Accuracy - How close you are(measured value) to your target(accepted value) THE MOLE - The mole is defined as a unit of measurement to measure the amount of matter in a substance - i.e. 16.1 mole of Carbon; means that the amount of carbon in the sample is 16.1 mol - 1 mol of substance contains (6.022)(10 23 ) particle (atoms, molecules, etc.) - (6.022)(10 23 ) is a special number, called Avogadro’s Number (Named after the man who discovered it - Examples ● 1 mol of Al contains (6.022)(10 23 ) atoms ● 1 mol of S 8 contains (6.022)(10 23 ) molecules ● 1 mol of C 12 h 24 0 11 contains (6.022)(10 23 ) molecules ● 1 mole of NaCl contains (6.022)(10 23 ) Formula Units - The amount of matter in moles is directly related to the number of particles( atoms, molecules, etc.) in a sample of matter - Therefore we gather this equation: ● Number of moles= (Number of atoms, molecules, or Formula Units)/(Avogadro’s number) ● This is shortened to n=N/N A ● “n” is number of moles in mol ● “N” is Number of atoms, molecules, or Formula Units ● “N A ” is Avogadro’s Number (6.022)(10 23 ) - Carbon-12, each C-12 has a mass of 12 amu (atomic units) - A sample of carbon has weighted average atomic mass of 12.011 amu - 1 mole of any substance has a mass in grams equal to its weighted average atomic mass (given on periodic table, don’t forget to calculate for total atomic mass of the substance by adding up all the atomic masses of each atom(s)!!! ) - Therefore we can we can gather another equation:

● Number of moles= (Entire mass of substance)/(Total Atomic mass of entire substance) ● Shortened to n=m/MM ● “n” is moles in mol ● “m” is mass in grams(g) ● “MM” is the total atomic mass of substance, also known as molar mass, is measured in grams per mole(g/mol), and is found by adding up all the atomic masses for each atom(s) in a molecule - Tip : From n=N/N A and n=m/MM we can derive the formula N/N A = m/MM PERCENT COMPOSITION - Law of Definite proportions: the element in chemical compounds are always present in the same proportion, by mass - Percent composition are ratios by the masses of an element in a compound , this is represented by the formula: (mass of atom(s))/(Total mass of the substance) - In a chemical equation, the ratios are mole to mole - i.e. 1NaCl —> 1Na + 1Cl, assume one mole of NaCl and find the percent composition of Cl ● 1 mol of NaCl ● 1 mol of Na ● 1 mol of Cl ● Therefore m Cl = (1 mol of Cl)( MM Cl ) and m Na = (1 mol of Na)(MM Na ) ● m Cl = 22.989768 g and m Na = 35.4527 g ● Therefore % Cl = (mCl)/(mNa + MCl) * 100% ● % Cl = 60.689252% —— don’t forget sig figs —–à = 60.6892% THE EMPIRICAL FORMULA - The simplest ratio of elements for a substance - Ie: CH 2 0; simplest ratio of elements for C 6 H 12 O 6 (Glucose) - It is found by finding the number of moles of each individual atom in a substance then dividing each of these by the smallest amount. These numbers are representative of the number of atoms there are in a single molecule orion for that atom. - Sample Question: A sample is found to contain 5.0 g of copper and 1.3 g of oxygen, and no other elements. Determine the empirical formula of this compound. ● What do we need? n Cu = ? and n O = ? ● n Cu = m Cu /MM Cu and n O = m O /MM O ● n Cu = 5.0g/(63.546g/mol) and n O = 1.3g/(15.9994g/mol) ● n Cu = 0.078683 mol and n O = 0.081253 mol

● Divide by small therefore n Cu / n Cu = 1 and n O / n Cu = 1.032667= 1 (this is because we cannot have decimals in a chemical formula) ● Since they are both 1 therefore the ratio of Cu to O is 1:1, therefore the Empirical Formula is Cu 1 O 1 or CuO