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Chapter 3 Stoichiometry

Jino Park Chris Yang. Chapter 3 Stoichiometry. 3.1 Counting by Weight. Average mass=total mass / number of objects ex) each jelly bean is 5g, you want 100 jelly beans. 3.2 Atomic Masses. The modern system of atomic masses is based on the 12 C as the standard.

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Chapter 3 Stoichiometry

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  1. Jino Park Chris Yang Chapter 3Stoichiometry

  2. 3.1 Counting by Weight • Average mass=total mass / number of objects • ex) each jelly bean is 5g, you want 100 jelly beans

  3. 3.2 Atomic Masses • The modern system of atomic masses is based on the 12C as the standard. • Carbon’s atomic mass is 12.01 g because it takes the average of the masses of its isotopes. • Basically, the atomic masses of the elements are just averages of their isotopes.

  4. 3.3 The Mole • 1 mole=6.02 x 1023 molecules • Dozeneggsmolemolecules • Use (6.02 x 1023 molecules / 1 mol) when converting

  5. 3.4 Molar Mass • The mass (g) of one mole of the compound. • Found by summing up the masses of each atom in a component.

  6. 3.6 Percent Composition of Compounds • Obtained by taking the ratio of the mass of each element in 1 mole of the compound to the total mass of 1 mole of the compound. • (mass of element in 1 mole/total mass in 1 mole) x 100 = percent composition

  7. 3.7 Determining the Formula of a Compound • Steps to find empirical formula: 1. Find % mass of each element in the compound 2. Assume there are 100g the compound, so each element would have the same mass (g) as its % mass 3. Convert each mass to moles; divide by the atomic mass 4. Divide all the moles by the lowest to find the ratio between the elements 5. The numbers should be very close to whole numbers; if not, multiply through (usually by 2 or 3) to get whole number ratios *If the molar mass is known, the molecular formula can be found • Ex) If a compound contains 71.65% Cl, 24.27% C, and 4.07% H by mass, and the molar mass is 98.96g/mol, what is the molecular formula of the compound?

  8. 3.8 Chemical Equations • A chemical reaction is the rearrangement of the atoms in one or more substances. • Represented by a chemical equation with reactants on the left side and products on the right. • Atoms are neither created nor destroyed • all atoms in reactant must be present in product • mass is conserved • Solid (s), liquid (l), gas (g), aqueous (aq)

  9. 3.9 Balancing Equations • Change only the coefficient, NOT the subscripts or atoms • The total numbers of an atom must be equal on both reactants and products • ex)C2H5OH + O2 CO2 + H2O

  10. 3.10 Stoichiometric Calculations: Amounts of Reactants and Products • Coefficients in the equation represent numbers of molecules. • Construct a mole ratio from the balanced equation. Ex. C3H8 + 5O2 3CO2 + 4H2O; (5 moles of O2/1 mol of C3H8) • Calculating Masses of Reactants & Products • Balance the equation • Convert mass of reactant/product to moles • Set up mole ratios using balanced equation • Use mole ratios to calculate number of moles of reactant/product • Convert from moles back to grams if required

  11. 3.11 Limiting Reagent • The reactant that runs out first, limiting the amounts of products • Use the limiting reagent to calculate the products • Determining a limiting reagent • Calculate the moles of each reactant • Compare ratio to the given equation • Percent Yield=actual/theoretical x 100%

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