Chapter 5 :Stoichiometry

Actual Yield

is the amount of product actually obtained from a reaction. % Yield = (Actual Yield/ Theoretical Yield)*100%

Theoretical Yield

is the amount of product that would result if all the limiting reagent reacted. *** This is the MAXIMUM obtainable result

Molar mass

is the mass of 1 mole of_______ in grams

Percent composition

of an element in a compound =((n x molar mass of \ element )/(molar mass of compound ))*100 n is the number of moles of the element in 1 mole of the compound

Stoichiometric Amounts:

Exact proportions indicated by the balanced chemical equation.

Molecular mass

(or molecular weight) is the sum of the atomic masses (in amu) in a molecule.

Grams to Moles to Moles to Grams...

1. Write balanced chemical equation 2.Convert quantities of known substances into moles 3.Use coefficients in balanced equation to calculate the number of moles of the sought quantity 4.Convert moles of sought quantity into desired units

Balancing Chemical Equations

1. Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation. 2. Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts. 3. Start by balancing those elements that appear in only one reactant and one product. 4.Balance those elements that appear in two or more reactants or products. 5. Check to make sure that you have the same number of each type of atom on both sides of the equation.

chemical reaction

A process in which one or more substances is changed into one or more new substances.

The Mole (mol):

A unit to count numbers of particles

How many hydrogen atoms are present in 25.6 g of urea [(NH2)2CO], which is used as a fertilizer, in animal feed, and in the manufacture of polymers? The molar mass of urea is 60.06 g.

How many hydrogen atoms are present in 25.6 g of urea [(NH2)2CO], which is used as a fertilizer, in animal feed, and in the manufacture of polymers? The molar mass of urea is 60.06 g. 25.6 g(NH2)2CO*(1 mol (NH2)2CO/60.06 g (NH2)2CO)*(4mol H/1 mol (NH2)2CO)*(6.022*10^23 H atoms/ 1 mol H)= 1.03*10^24 H atoms

Limiting Reagent:

Reactant used up first in the reaction. 2NO + O2 ---- 2NO2 NO is the limiting reagent O2 is the excess reagent

Phosphoric acid (H3PO4) is a colorless, syrupy liquid used in detergents, fertilizers, toothpastes, and in carbonated beverages for a "tangy" flavor. Calculate the percent composition by mass of H, P, and O in this compound.

Solution The molar mass of H3PO4 is 97.99 g. The percent by mass of each of the elements in H3PO4 is calculated as follows: %H=(3(1.008 g) H/ 97.99 g H3PO4)*100%= 3.086% %P=(30.97 g P/ 97.99 g H3PO4)*100%= 31.61% %O=(4(16.00 g)O/ 97.99 g H3PO4)*100%=65.31% Check Do the percentages add to 100 percent? The sum of the percentages is (3.086% + 31.61% + 65.31%) = 100.01%. The small discrepancy from 100 percent is due to the way we rounded off.

Calculate the molecular masses (in amu) of the following compounds: sulfur dioxide (SO2), a gas that is responsible for acid rain caffeine (C8H10N4O2), a stimulant present in tea, coffee, and cola beverages

Solution To calculate molecular mass, we need to sum all the atomic masses in the molecule. For each element, we multiply the atomic mass of the element by the number of atoms of that element in the molecule. We find atomic masses in the periodic table (inside front cover). a)There are two O atoms and one S atom in SO2, so that molecular mass of SO2 = 32.07 amu + 2(16.00 amu) = 64.07 amu b)There are eight C atoms, ten H atoms, four N atoms, and two O atoms in caffeine, so the molecular mass of C8H10N4O2 is given by: 8(12.01 amu) + 10(1.008 amu) + 4(14.01 amu) + 2(16.00 amu) = 194.20 amu

Sulfur (S) is a nonmetallic element that is present in coal. When coal is burned, sulfur is converted to sulfur dioxide and eventually to sulfuric acid that gives rise to the acid rain phenomenon. How many atoms are in 16.3 g of S?

We need two conversions: first from grams to moles and then from moles to number of particles (atoms). Because 1 mol S = 32.07 g S the conversion factor is 1 mol S/32.07 g S Avogadro's number is the key to the second step. We have 1 mol = 6.022 × 1023 particles (atoms) (16.3 g S) *(1 mol S/32.07 g S)* (6.022*10^23 S atom/1 mol S)= answer= 3.06*10^23

Balance the following chemical equations: ___ C7H16 + ___ O2 ----- ___ CO2 + ___ H2O ___ Ca(OH)2 + ___ H3PO4 ----- ___ Ca3(PO4)2 + ___ H2O ___ CH4 + ___ Br2 ---- ___ CBr4 + ___ HBr ___ I2 + ___ XeF2 ----- ___ IF3 + ___ Xe ___ FeBr3 + ___ NH4OH ----- ___ Fe(OH)3 + ___ NH4Br

___ C7H16 + _11_ O2 ----- _7_ CO2 + _8_ H2O _3_ Ca(OH)2 + _2_ H3PO4 ----- ___ Ca3(PO4)2 + _6_ H2O ___ CH4 + _4_ Br2 ----- ___ CBr4 + _4_ HBr ___ I2 + _3_ XeF2 ----- _2_ IF3 + _3_ Xe ___ FeBr3 + _3_ NH4OH ------ ___ Fe(OH)3 + _3_ NH4Br