Stoichiometry 1 PDF

**[Stoichiometry 1]** Stoichiometry is the branch of chemistry that tells us the relationship between the amounts of reactants and products in a chemical reaction. Chemists need a good knowledge of stoichiometry because it enables them to decide on the quantities of reactants needed to form a certa...

**[Stoichiometry 1]** Stoichiometry is the branch of chemistry that tells us the relationship between the amounts of reactants and products in a chemical reaction. Chemists need a good knowledge of stoichiometry because it enables them to decide on the quantities of reactants needed to form a certain amount of product. **[To Calculate the Percentage of each Element in a Compound]** Calculate the percentage composition by mass of each element in calcium pxide, CaO. *Calculate the relative molecular mass of CaO.\ *Ca = 40\ [O = 16\ ]Rel. Molecular Mass = 56 Percentage of Ca = [\$\\frac{40}{56}\$]{.math.inline} x 100 = 71.43% Percentage of O = [\$\\frac{16}{56}\$]{.math.inline} x 100 = 28.57% The percentage composition of CaO by mass is 71.43% Ca and 28.57% O. **[To Calculate the Empirical Formula of a Compound Given the Percentage Composition]** \*\*Definitions: The molecular formula of a compound is a formula which shows the number and type of each atom present in a molecule of that compound.\ The empirical formula of a compound is the formula showing the simplest whole number ratio of the numbers of different atoms present in the molecule.\*\* Molecular formula = empirical formula x n (where n is a whole number) +-----------------------------------------------------------------------+ | The empirical formula of glucose is CH~2~O. When the relative | | molecular mass of glucose is measured by a mass spectrometer, it is | | found to be 180. What is the molecular formula of glucose?\ | | Molecular formula = empirical formula x n\ | | Molecular formula = (CH~2~O) x n\ | | 180 = (12 + 2 + 16) x n\ | | 180 = 30 x n\ | | n = 6 | | | | (CH~2~O) x 6 = | | | | C~6~H~12~O~6~ | +-----------------------------------------------------------------------+ **[To calculate the Empirical Formula of a Compound Given the Masses of Reactants and Products]** In an experiment to determine the formula of an oxide of copper, hydrogen gas was passed over the heated oxide. It was found that 5.4 g of the oxide gave 4.8 g of copper. Calculate the empirical formula of the oxide. Copper oxide + hydrogen copper\ 5.4 g 4.8 g Mass of copper oxide = 5.4 g\ Mass of copper in copper oxide = 4.8 g\ Therefore, mass of oxygen in copper oxide = 5.4 -- 4.8 = 0.6 g No. moles of copper atoms in copper oxide = [\$\\frac{4.8}{63.5}\$]{.math.inline} = 0.076 moles No. moles of oxygen atoms in copper oxide = [\$\\frac{0.6}{16}\$]{.math.inline} = 0.038 moles\ Ratio of Cu:O = 0.076:0.038 = 2:1\ Empirical formula of copper oxide is Cu~2~O Thus, the following reaction has occurred\ Cu~2~O + H~2~ 2Cu + H~2~O **[Calculating Masses of Reactants or Products from Balanced Chemical Equations]** Magnesium reacts with oxygen to produce magnesium oxide according to the equation 2Mg + O~2~ 2MgO\ If a student burns 9 g of magnesium in excess oxygen, what mass of magnesium oxide will be formed? 2Mg + O~2~ 2 MgO 2Mg 2MgO\ 2 moles 2 moles\ [2(24) g] [ 2 (24 + 16) g]\ 48 g Mg 80g MgO Therefore, 1 g Mg = [\$\\frac{80}{48}\$]{.math.inline} g MgO 9g Mg [\$\\frac{9\\ x\\ 80\\ }{58}\$]{.math.inline} g MgO 15 g of magnesium oxide will be formed from 9 g of Mg. **[Calculating Volumes of Gases from Balanced Chemical Equations]** Carbon dioxide is formed when carbon is burned in air according to the equation.\ C + O CO~2~\ What volume of carbon dioxide is formed when 30 g of carbon are burned in excess oxygen? C CO~2~\ 1 mole 1 mole\ 12 g C 22.4 L of CO~2~ at s.t.p\ 1 g C [\$\\frac{22.4}{12}\$]{.math.inline} L of CO~2~\ 30 g C [\$\\frac{30\\ x\\ 22.4}{12}\$]{.math.inline} L of CO~2~ Answer: 56L of Carbon dioxide