Formula & Equation Stoichiometry

What is stoichiometry? Deals with the specifics of QUANTITY in chemical formula or chemical reaction.

Review: Counting Atoms Remember subscripts tell you how many atoms… … Coefficients tell you how many TOTAL molecules compounds

How many oxygen atoms in each? NH4NO3 C8H8O4 O3 C3H5(NO3)3 (3) (4) (3) (9)

Counting Atoms Practice Ditto Example #1: (NH4)2CO3 * First list the types of atoms and then count each. N: H: C: O: 2 8 1 3

Counting Atoms Practice Ditto cont. Example #2: CaCrO4 * First list the types of atoms and then count each. Ca: Cr: O: 1 4

Counting Atoms Practice Ditto cont. Example #3: Ca3(PO4)2 * First list the types of atoms and then count each. Ca: P: O: 3 2 8 **Do the rest on your own!**

Atomic Mass Mass of an atom of one element based on percent abundances and masses of all isotopes (on PT) Units: amu

Molecular Mass Mass of a molecular compound (ex: H2O, CO2) Units: amu

Formula Mass Mass of an ionic compound (ex: NaCl, MgI2) Units: amu

Gram Formula Mass Mass expressed in grams

Gram Molecular Mass = H2O O + H + H + + Molecular Mass = 15.9994 amu + 1.00794 amu +1.00794 amu Gram Molecular Mass = 18.0 g

How do I calculate molecular or formula mass? First: Identify and count atoms in the compound. Second: Locate atomic mass of each element. Third: Multiply mass x number of atoms and total all elements. Voila!

Molecular Mass Example (Covalent Compound) Example: H2O H- 2 x 1 O- 1 x 16 Molecular Mass = 18 g

Formula Mass Example (ionic compound) Example: NaCl Na- 1 x 23 Cl- 1 x 35.5 Formula Mass = 58.5 g

Formula Mass Practice

Percent Composition by Mass Experimentally- Use masses given in problem Part x100 Whole Mass of Element x 100 Mass of Compound

Example: A compound containing carbon and hydrogen has a mass of 16 grams. When decomposed, 12.0 grams are found to be carbon. What is the percent by mass of carbon in the compound?

Percent Composition by Atomic Mass Theoretically- Use Atomic Masses Formula Part x 100 Whole (#atoms of element) x (atomic mass) x100 Formula mass of compound

Hydrated Crystals Some ionic compounds are found to have surrounding water molecules. These compounds are called Hydrated Crystals. The percent water hydration can be found. Example CuSO4 . 5H2O The dot shows that 5 H2O molecules are attached to 1 CuSO4 molecule.

Percent Water Hydration % Hydration = Total mass of Water x 100 Formula Mass CuSO4 . 5H2O Cu – 1 x 63.5 = 63.5 S – 1 x 32 = 32 O – 4 x 16 = 64 H2O – 5 x 18 = 90 Formula Mass = 249.5 % Hydration = (90/249.5) x 100 = 36% H2O

% Hydration Example Problems MgCl2 . 6H2O CaCl2 . 2H2O Ba(OH)2 . 8H2O CoSO4 . 7H2O NH4Al(SO4)2 . 12H2O

The Mole Avagadro’s Number is the number of atoms (for monatomic elements) or molecules (for diatomic elements & compounds) in 1 mole of any substance. Avagadro’s Number is 6.02 x 1023 That means that in 1 mole of any sample of matter there are 6.02x1023 particles. (atoms or molecules depending on the substance)

The Mole There are 6.02 x 1023 particles in one mole of any substance (element or compound). One particle can be one atom or one molecule. That’s a BIG number!!!!! Even though for any substance the number of particles in 1 mole is always the same different substances have different masses so the mass of 1 mole will be different for all substances. Atomic / Formula / Molecular Mass = Molar Mass because it is the mass of 1 mole of any substance. Example: 1 mole of H2O = 18g

Mole Relationships or Equalities 1 mole = gram formula mass (molecule) 1 mole = atomic mass (element) 1 mole = 6.02 x 1023 particles 1 mole = 22.4L (for gases) Formula mass = 6.02x1023 particles 22.4L of a gas = 6.02x1023 particles Formula mass = 22.4L of a gas

Table T : Mole Calculations For questions that involve mass (gram) to mole conversions use the “Mole Calculation” formula on Table T. In order to use this formula you will need to plug in what is given and solve for the unknown value. If solving for moles then plug in given mass and gram formula mass (GFM) If solving for a mass then plug in moles and GFM and solve for given mass You will usually have to calculate the gram formula mass of the substance unless it is given. See Table T

Examples Example #1: What is the mass of 1.75 moles of oxygen gas (Hint: oxygen is diatomic)? Convert from moles to grams using the mole equation on Table T by solving for given mass # of moles = given mass gram-formula mass x 32 1st find the gram formula mass (GFM)of oxygen remember oxygen is a diatomic element (O2). O: 2 x 16 g= 32 g 1.75 = X= 56 g

Examples Example #2: How many moles are in 42 g of water? 1st find the GFM of H2O. H: 2 x 1 = 2 O: 1 x 16 = 16 2 + 16 = _18 g_ 1 mol Use mole calculation equation solve for # of moles # of moles = given mass gram-formula mass 42 18 X = 2.3 mol

Examples Example #3: How many moles of potassium chromate are in a 500 g sample? 1st find the molecular mass of potassium dichromate (K2CrO4). K:2 x 39 = 78 Cr: 1 x 52 = 52 O: 4 x 16 = 64 78+104+112 = _194 g_ 1 mol Use mole calculation equation solve for moles 500 194 X = 2.6 mol

Example 4 How much does 2.0 moles of carbon dioxide weigh?

Example 5 What is the mass of 0.50 moles of sulfur dioxide at STP?

Example 6 How many moles of helium gas are in 20.0 grams of helium? Try the Practice problems

Empirical & Molecular Formulas Empirical Formula: formula that contains the lowest whole number ratio of atoms. Molecular Formula: the actual formula for a compound (does not have to be empirical but it can) The subscripts in any formula represents the number of moles of atoms in a 1 mole sample. The subscripts in an empirical formula can tell you the mole ratio of atoms in a compound. Example: In 1 mole of H2O there are 2 moles of Hydrogen atoms and 1 mole of oxygen atoms.

Calculating the Molecular Formula Given Formula Mass and Empirical Formula Since the empirical formula is the lowest mole ratio of atoms in a compound the molecular formula mass must be a multiple of the empirical formula mass.

Step 1: Find the empirical formula mass. Example 1: What is the molecular formula for a compound that has an empirical formula of CH2O and the molecular mass is 180g. Step 1: Find the empirical formula mass. Step 2: Divide molecular mass given by the empirical mass you calculated in step 1. Step 3: Multiply all of the subscripts in the empirical formula by the multiple you calculated in step 2

Example 1 Step 1: CH2O Empirical Mass C – 1 x 12 = 12 H – 2 x 1 = 2 O – 1 x 16 = 16 30 Step 2: Calculate Multiple (Formula mass/Empirical mass) 180/30 = 6 Step 3: Multiply Empirical Formula by Multiple 6(CH2O) = C6H12O6

Example 2: A mass of a given compound is 9.2g. It is determined that 2.8g is nitrogen and 6.4g is oxygen. What is the empirical formula of this compound? Step 1: Convert the given mass of each element to moles by dividing by the atomic mass. Step 2: Plug moles values from step 1 into the formula and divide both by the lowest to get a whole number ratio. If ratio is still not whole number then multiply the whole formula by 2.

N: 2.8/14 = .2mol. O: 6.4/16 = .4mol. N.2/.2 O.4/.2 Answer = NO2 Example 2: A mass of a given compound is 9.2g. It is determined that 2.8g is nitrogen and 6.4g is oxygen. What is the empirical formula of this compound? N: 2.8/14 = .2mol. O: 6.4/16 = .4mol. N.2/.2 O.4/.2 Answer = NO2

Example 3: A compound is found to be 25. 9% nitrogen and 74. 1% oxygen Example 3: A compound is found to be 25.9% nitrogen and 74.1% oxygen. What is the emeprical formula of this compound? Step 1: Treat the percentage values as mass values (assume sample size is 100g) and convert to moles. Step 2: Plug mole values into the equation and divide both by the lowest to get whole number ratios. If ratio is still not whole number then multiply the entire formula by 2.

Example 3: A compound is found to be 25. 9% nitrogen and 74. 1% oxygen Example 3: A compound is found to be 25.9% nitrogen and 74.1% oxygen. What is the emeprical formula of this compound? N: 25.9/14 = 1.85mol. O: 74.1/16 = 4.63mol. N1.85/1.85O4.63/1.85 = 2(NO2.5) Answer = N2O5

Mole Relationships in Balanced Equations Balanced chemical equations tell us the number and type of substances involved in a chemical reaction. The coefficients indicate the amounts of reactants that go into a reaction and the amount of product formed. In chemistry we also use the coefficients in a balanced equation to identify the mole ratios of reactants and products.

Mole Ratios in a Balanced Equation 2H2 + O2 → 2H2O In the balanced equation for the synthesis of water it shows that for every 2 hydrogen molecules you need 1 oxygen molecule and 2 water molecules are produced. It can also be said that for every 2 moles of hydrogen and 1 mole of oxygen 2 moles of water are produced. The coefficient can be read as molecules or moles, in lab we use moles because it is a measureable amount.

Solving Equation Stoichiometry Problems Remember that balanced equations are mole ratios of reactants and products. Ex 1: Synthesis of water 2H2 + O2 → 2H2O Means 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water. The Ratio of reactants and products are 2:1:2. We can use the molar ratios from a balanced equation to determine the amounts of reactants used or products produced with varying amounts.

Examples Problem 2H2 + O2 → 2H2O Ex 2: How many moles of water are Examples Problem 2H2 + O2 → 2H2O Ex 2: How many moles of water are produced if 2 moles of oxygen reacted? Set up a proportion Solve for unknown

How many moles of carbon dioxide will be produced by the complete combustion of 2.0 moles of glucose (C6H12O6) First: write the equation and balance it. C6H12O6 + 6O2 ⟶ 6CO2 + 6H2O Next: set up a proportion Finally: solve for the number of moles

How many moles of ammonia can be produced from 9 How many moles of ammonia can be produced from 9.0 moles of hydrogen reacting with nitrogen? Try the practice!!!