Stoichiometry Notes. Equations to MEMORIZE 1 mol = (molar mass) g – Get it from the atomic mass. Add it up if you have a molecule. (O=15.999, so O 2 =31.998)

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Presentation transcript:

Stoichiometry Notes

Equations to MEMORIZE 1 mol = (molar mass) g – Get it from the atomic mass. Add it up if you have a molecule. (O=15.999, so O 2 =31.998) 1 mol = x atoms or molecules 1 mol = 22.4 L (of a gas) Avagadro’s Number

Congruent Triangles 1 mol H 2 O 3.5 mol H 2 O g H 2 O X g H 2 O Question Amount Equation Amount Question Amount Equation Amount = What is the mass in grams of 3.5 moles of water ? Molar mass = mass Of 2 H’s and 1 O __X g H 2 O__ 3.5 mol H 2 O 1 mol H 2 O 3.5 mol H 2 O g H 2 O = 1 x X = 3.5 x X=63.05 g H 2 O ______ 2 ___

Cross Multiplication Review A x Y = Z x B A Z B Y =  3 x Y = 4 x Y =  Y = 4 x x Y = 4 x Y = 6.67

Basic conversions [Question amount/Equation amount = Question amount/Equation amount] Mole  Grams – 2.5 moles of O 2 = how many grams? 1 mole O 2 = g O 2 (the weight of 2 Oxygen) 2.5 mol O 2 = __X g O 2 __ – 8 g of O 2 = how many moles __8 g O 2 = X mol O 2 1 mol O g O g O 2 1 mol O 2 X = g O 2 X =.25 mol O 2 Question Amount Equation Amount Question Amount Equation Amount =

Basic conversions [Question amount/Equation amount = Question amount/Equation amount] Moles  Atoms (or molecules) – 7.3 moles of H 2 O = how many molecules? 1 mole H 2 O = x molecules H 2 O – (This is Avagadro’s Number) 7.3 mol H 2 O = ____X Molecules H 2 O____ – 3.92 x molecules of H 2 O = how many Moles 3.92x10 23 Molecules H 2 O = X mol H 2 O 1 mol H 2 O 6.022x10 23 Molecules H 2 O 6.022x10 23 Molecules H 2 O 1 mol H 2 O X = 4.40 x molecules H 2 O X =.65 mol H 2 O Question Amount Equation Amount Question Amount Equation Amount =

2 step conversion Atoms  Grams (Have to convert to moles first) – 3 g of C 4 H 10 = how many molecules? __3 g C 4 H 10 __ = X mol C 4 H 10 X =.05 mol C 4 H mol C 4 H 10 = X molecules C 4 H 10 ____ – x molecules of C 4 H 10 = how many g? 3.058x10 24 molecules C 4 H 10 = X mol C 4 H x10 23 molecules C 4 H 10 1 mol C 4 H 10 – X = 5.08 mol C 4 H mol C 4 H 10 = X g C 4 H 10 1 mol C 4 H g C 4 H 10 1 mol 6.022x10 23 molecules C 4 H g C 4 H 10 1 mol C 4 H 10 X = 3.01 x molecules C 4 H 10 X = g C 4 H 10

Stoichiometry Moles to moles 2 C 4 H O 2  8 CO H 2 O – How many moles H 2 O will 3.5 moles of C 4 H 10 make? 3.5 mol C 4 H 10 = X mol H 2 O – Question amount/equation amount = question amount/equation amount 2 mol C 4 H mol H 2 O X = 17.5 mol H 2 O

Moles to grams 2 C 4 H O 2  8 CO H 2 O How many moles of CO 2 will 2.4 grams of C 4 H 10 make? Same as before but have to substitute in the mass of 2 moles 2.4 g C 4 H 10 = X mol CO 2 2 mol C 4 H 10 8 mol CO 2 2 x (4(12.011) + 2(1.008)) = {2 moles x 4 Carbons + 2 Hydrogens} 2.4 g C 4 H 10 = X mol CO g C 4 H 10 8 mol CO 2 X =.17 mol CO 2

Moles to grams – 2 C 4 H O 2  8 CO H 2 O – How many grams of CO 2 will 2.4 moles of C 4 H 10 make? 2.4 mol C 4 H 10 = _X g CO 2 _ 2 mol C 4 H 10 8 mol CO 2 8 x (1(12.011) + 2(15.999)) = mol C 4 H 10 = X g CO 2 2 mol C 4 H g CO 2 X = g CO 2

Grams to grams 2 C 4 H O 2  8 CO H 2 O – 17 g of Oxygen will produce how many grams of Carbon Dioxide? – Have to substitute on both ends _17 g O 2 _ = X mol CO 2 13 mol O 2 8 mol CO 2 13 x 2(15.999) = x (1(12.011) + 2(15.999)) = g O 2 = X g CO g O g CO 2 X = g CO 2

Volume to moles 2 C 4 H O 2  8 CO H 2 O – How many liters of CO 2 gas are produced by 5 mol of C 4 H 10 ? – Good luck measuring the weight of a gas! Gasses are usually dealt with in volume. – 22.4 L of any gas (at standard pressure) = 1 mol of that substance. 5 mol C 4 H 10 = _X L CO 2 _ 2 mol C 4 H 10 8 mol CO 2 8 x 22.4 = mol C 4 H 10 = X L CO 2 2 mol C 4 H 10 = L CO 2 (Moles to volume would be the same but the left side would have the substitution) X = 448 L CO 2

Volume to grams 2 C 4 H O 2  8 CO H 2 O How many Liters of O 2 do you need to produce 105 g of H 2 O? Same as before but you will have two substitutions. – 105 g H 2 O = X L O 2 10 mol H 2 O = 13 Mol O 2 – 10 x (2(1.008) + 1(15.999) ) = – 13 x 22.4 = – 105 g H 2 O = X L O g H 2 O L O 2 X = L CO 2

Limiting Reagents = what you run out of first 2 C 4 H O 2  8 CO H 2 O If you react 80 g of C 4 H 10 with 15 g of O 2, which is the limiting reagent? Question/Equation. Then compare the numbers 80 =.69 (2 x (4(12.011) + 10(1.008))) 15 =.04 (13 x 2(15.999)) Lowest answer is the Limiting Reagent! If you want to know how much product you will make, you need to use the numbers of the limiting reagent. You will end up with extra of the other one that does not react.