Unit 4: The MOLE & STOICHIOMETRY. Essential Question: What is a MOLE and why do we use it to measure chemical things PRECISELY?

Transcription

1 Unit 4: The MOLE & STOICHIOMETRY Essential Question: What is a MOLE and why do we use it to measure chemical things PRECISELY?

2 What makes a figure SIGNIFICANT? n Non-Zero DIGITS n n n 3600 n FOUR n THREE n TWO n Sandwiched zeroes n n n 10.5 n FOUR n FIVE n THREE

3 What makes a figure SIGNIFICANT? n Zeros AFTER sig figs n n THREE n DIGITS before all non significant zeroes n 360 n 36,000 n TWO n TWO

4 Sig Figs in mathematical equations: n Addition (or Subtraction) à ROUND FIRST, then compute 212 g g g = 216 g n Multiplication (or Division) à compute, THEN ROUND 4.16 cm x 2.2 cm x 2.00 cm = 18 cm 3 *factors with the least #of sig figs determines the amount of sig figs in the answer

5 Units Conversion: n TABLE C! prefixes/scientific notation n conversion factors of base 10 n *remember, use calculator, x10^ is the same as 2 nd function, EE n 10 3 = kilo- n 10-1 = deci- n 10-2 = centi n 10-3 = milli- n 10-6 = micro- n 10-9 = nano- n = pico-

6 Units Conversion: n TABLE D! symbols, type of SI unit n (le système internationale d Unités) n Basic Selected Units: n Length = meter n mass = kilogram n temp = kelvin à table A (C +273) n time = second n amount of substance = mole n electric current = ampere

7 Units Conversion: + How to convert (dimensional analysis): *look at do-now problems take a minute, respond to the question on your notes sheet. *IT S JUST LIKE SIMPLE PROPORTIONS IN ALGEBRA! Let s try! (**ALWAYS show units = be precise ) n How many quarters do I have in $120?

8 Units Conversion: + Practice Problem #2: How many Kilograms are in 2500 grams? we know this! plug in factors cross out units that CANCEL compute and solve!

9 Formula Mass: n What are the atomic masses of K? C? O? n Identify how many atoms of each element are present in: K 2 CO 3 n How do we find the formula mass of a compound then? n Ex: K 2 CO 3 à Potassium Carbonate n #K s = 39.1amu each (39.1x2 = 78.2) n #C s = 12.0 amu n #O s = 16.0 amu each (16x3 = 48.0) FORMULA MASS of K 2 CO 3 = amu because = amu

10 GRAM - Formula Mass: n Gram-Formula Mass: (exactly what it sounds like!) is the formula mass of a substance measured in GRAMS instead of AMU n Ex) we just calculated K 2 CO 3 s formula mass it was amu. n in GFM, we just convert the units to GRAMS instead, so the GFM or Molar Mass of K 2 CO 3 is grams

11 Percent Composition: n Just like we can customize our pizza orders with various PERCENTS or portions of toppings, we can understand the relationship between how much mass each component, or TOPPING, of a chemical compound, or PIZZA is contributing to the compound s TOTAL MASS n.yum!

12 Percent Composition: THEREFORE Percent Composition of a substance represents the composition as a percentage of each element compared with the total mass of the compound. Ex: What is the percentage of oxygen in potassium chlorate (KClO 3 )? à Determine formula mass: K = 1 x 39.1 amu = 39.1 amu Cl = 1 x 35.5 amu = 35.5 amu O = 3 x 16.0 amu = 48.0 amu TOTAL = amu. THEN à

13 Percent Composition: (now that it s calculated.) K = 1 x 39.1 amu = 39.1 amu Cl = 1 x 35.5 amu = 35.5 amu O = 3 x 16.0 amu = 48.0 amu TOTAL = amu à Now calculate the % of Oxygen in the whole compound: Oxygen in compound = 48.0 amu total = amu so, 48.0/122.6 = x 100% = 39.2% **look at table T** last page reference table!

14 THE MOLE Avogadro s # : One Dozen = 12 ONE MOLE = x particles Amedeo Avogadro: Scientist responsible for quantifying a MOLE to be the # of particles that constitutes a substances gram formula mass. Abbreviation = mol

15 THE MOLE Avogadro s # : so if 1 MOLE = x particles, then How many particles are in 2 moles? à x particles How many particles are in moles? à x particles How many MOLES are in x particles? à moles

16 THE MOLE Avogadro s # : Practice Problem #1: solving for grams How many grams are present in 40.5 mol of sulfuric acid (H 2 SO 4 )? a) Solve for the GFM (molar mass): H = 2 x 1.0 amu = 2 amu S = 1 x 32.1 amu = 32.1 amu O = 4 x 16.0 amu = 64.0 amu à 98.1 amu = g SO molar mass (mm) of H 2 SO 4 = 98.1 g b) Convert # of moles to grams how? *** à let s see!

17 THE MOLE Avogadro s # : + *molar mass (mm) of H 2 SO 4 = 98.1 g à this mass is the mass of 1 MOLE of H 2 SO 4 G MM M

18 THE MOLE Avogadro s # : + *molar mass (mm) of H 2 SO 4 = 98.1 g grams = UNKNOWN à *cover it! mm = 98.1 g moles = 40.5 mol SOLVE: (perform operation on Δ) mm (98.1 g) x mol (40.5) = 3970 grams H 2 SO 4

19 THE MOLE Avogadro s # : Practice Problem #2: solving for grams moles of NaCl are equivalent to how many grams? a) Solve for the GFM (molar mass): Na = 1 x 23.0 amu = 23.0 amu Cl = 1 x 35.5 amu = 35.5 amu à 58.5 amu= g SO molar mass (mm) of NaCl = 58.5 g b) Convert # of moles to grams: *triangle mm (58.5) x mol (10.25) = grams NaCl

20 THE MOLE Avogadro s # : + Practice Problem #3: solving for moles How many moles are equivalent to 4.75 grams of sodium hydroxide (NaOH)? a) Solve for the GFM (molar mass): Na = 1 x 23.0 amu = 23.0 amu O = 1 x 16.0 amu = 16.0 amu H = 1 x 1.0 amu = 1.0 amuà 40.0 amu = g SO molar mass (mm) of NaOH = 40.0 g b) Convert # of grams to moles TRIANGLE! *cover à moles = unknown total g (4.75)/mm (40.0g) = mol (round) à mol NaOH

21 THE MOLE Avogadro s # : Practice Problem #4: solving for moles 1000 grams of ammonia, NH 3 equals how many moles? a) Solve for the GFM (molar mass): N = 1 x 14.0 amu = 14.0 amu H = 3 x 1.0 amu = 3.0 amuà 17.0 amu = g SO molar mass (mm) of NH 3 = 17.0 g b) Convert # of grams to moles TRIANGLE! *cover à moles = unknown total g (1000)/mm (17.0g) = 58.8 mol NH 3

22 THE MOLE in Balanced Equations *coefficients represent the MOLE RATIO of necessary amounts of reactants & products for the reaction to occur. 2C 2 H 6 + 7O 2 à 4CO 2 + 6H 2 O Moles C 2 H 6 Moles O 2 Moles CO 2 Moles H 2 O

23 THE MOLE in Balanced Equations Example problem: How many moles of water will be produced from the complete combustion of 3.0 mol of ethane according to the following equation? 2C 2 H 6 (g) + 7O 2 (g) à 4CO 2 (g) + 6H 2 O (g) Mol of C 2 H 6 = 3.0 mol Ratioà Moles ethane : Moles water 2 : 6 Proportion = 3.0 mol = x 2 (coefficient) 6 (coefficient) ANSWER: 2x = (3.0)(6). x = 9.0 mol H 2 O

24 THE MOLE in Balanced Equations Example Gases: How many liters of carbon dioxide gas will be produced from the complete combustion of 30.0 L of ethane according to the following equation? 2C 2 H 6 (g) + 7O 2 (g) à 4CO 2 (g) + 6H 2 O (g) Volume of C 2 H 6 = 30.0 liters Ratioà Moles C 2 H 6 : Moles CO 2 2 : 4 Proportion = 30.0 L = x 2 4 ANSWER: x = 60 L