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Unit I: Level 4MeasurementBefore dealing with the specifics of the rules for determining the significant figures in a calculated result, we need to be able to round numbers correctly. To round a number, first decide how many significant figures the number should have. Once you know that, round to that many digits, starting from the left. If the number immediately to the right of the last significant digit is less than 5, it is dropped and the value of the last significant digit remains the same. If the number immediately to the right of the last significant digit is greater than or equal to 5, the last significant digit is increased by 1.Consider the measurement 207.518 m. Right now, the measurement contains six significant figures. How would we successively round it to fewer and fewer significant figures? Follow the process below.Number of Significant FiguresRounded ValueReasoning6207.518All digits are significant5207.528 rounds the 1 up to 24207.52 is dropped32085 rounds the 7 up to 822108 is replaced by a 0 and rounds the 0 up to 1.0 upto 112001 is replaced by a 0Notice that the more rounding is done, the less reliable to figure is. An approximate value may be sufficient for some purposes, but scientific work requires a much higher level of detail.ReviewWhy do we round numbers?What do we need to know before we round a number?What is “rounding up”?What is “rounding down”?Significant FiguresSignificant digits are all digits known with certainty, plus one final digit that is uncertain or is estimated. Rules1. Zeros appearing between nonzero digits are significant.40.7 3 SD 87 009 5 SD2. Zeros appearing in front of nonzero digits are not significant.0.0958975 SD0.000009 1 SD3. Zeros at the end of a number and to the right of the decimal point are significant.85.00 4 SD9.000000000011 SD0.009000 4 SD4. Zeros at the end of a whole number without a decimal are not significant.2000 1 SD5. Zeros at the end of a whole number with a decimal are significant.2000. 4 SD6. When multiplying or dividing, the answer contains the same number of significant digits as the smaller (smallest) number in the problem.40. x 100 = 40002 SD 1 SD 1 SD0.00640 ÷ 0.80 = 0.00803 SD 2 SD 2 SDRules for Rounding OffIf that first digit to be dropped is less than 5 (that is, 1, 2, 3 or 4), drop it and all the digits to the right of it. If that first digit to be dropped is more than 5 (that is, 6, 7, 8 or 9), increase by 1 the number to be rounded, that is, the preceding figure (to the digit being dropped). If that first digit to be dropped is 5, round the digit that is to rounded off so that it will be even. Keep in mind that zero is even when rounding off. Practice ProblemsSignificant Figures and Rounding Practice ProblemsA. Determine the number of significant figures in each measurement1.) 340,438 g 2.) 3,000,002 cm3.) 87,000 ms 4.) 1.040 s 5.) 4080 kg 6.) 0.2080 mL 7.) 961,083,110 m 8.) 0.0000481 g 9.) 400,000 g 10.)20.00 L 11.)0.0030 mm 12.)1.2000 g 13.)32.4 0C 14.)41.05 kg 15.)0.000089 kg 16.)0.340 dL 17.)7301.00 g 18.)1.0 x 102 B. Write the following in three significant figures, rounding where necessary (you may have to convert to scientific notation)1.) 0.0030850 km 2.) 5808 3.) 3.0823 g 4.) 34.654 mg C. Provide the following information. Round where appropriate1.) Change 2346.2100 to show 5 sig figs2.) Change 60000 to show 6 sig figs3.) Change 0.00085066 to show 1 sig fig4.) Change 78.5 to show 4 sig figs5.) Change 25798124100436.005630 to show 3 sig figsD. Use rounding rules when you complete the following1.) 34.3 m / 35.8 m x 33.7 m2.) 0.056 kg x 0.0783 kg / 0.0323 kg3.) 309.1 mL / 158.02 mL x 238.1 mL4.) 1.03 mg x 2.58 mg x 4.385 mg5.) 8.376 km / 6.153 km6.) 34.24 s /12.4 s7.) 804.9 dm x 342.0 dm8.) (6.38 x 102) x (1.57 x 102)9.) 34.3 cm x 12 cm10.) 0.054 mm x 0.3804 mm11.) 45.1 km x 13.4 km12.) 45.5 g ÷ 15.5 mL13.) 35.43 g ÷ 24.84 mL14.) 0.0482 g ÷ 0.003146 mLThe Metric SystemThe first standardized system of measurement, based on the decimal, was proposed in France about 1670. However, it was not until 1791 that such a system was developed. It was called the "metric" system, based on the French word for measure. The driving force was the growing importance of weights in the sciences, especially chemistry. At that time, every country had their own system of weights and measures. England had three different systems just within its own borders!! On May 20, 1875, delegates of 17 countries signed the Meter Convention. It was amended in 1921, and today, 48 countries are signatories. The modern metric system has been renamed Systeme International d'Unites (International System of Units), and is denoted by the letters SI. SI was established in 1960, at the 11th General Conference on Weights and Measures. It was then that units, definitions, and symbols were revised and simplified. Parts of the Metric System There are three major parts to the metric system: the seven base units, the prefixes and derived units built from the base units. Here is a list of the base units that make up the metric system: Physical Quantity Name of SI unit Symbol for SI unit amount of substance mole mol electric current ampere A length metre (meter) m luminous intensity candela cd mass kilogram kg temperature Kelvin K time second s volume (non SI unit)literLvolume (SI unit) cubic meterm3YOU NEED TO KNOW THESE! Metric PrefixesA metric prefix is a modifier on the root word and it tells us the unit of measure. For example, centigram means we are count in steps of one one-hundredth of a gram, μg means we count by millionths of a gram. In order to properly convert from one metric unit to another, you learn them. You will also need to determine which of two prefixes represents a bigger amount AND you will also need to determine the exponential "distance" between two prefixes. A List of Common Metric PrefixesPrefix Symbol Numerical ValueExponential Value tera T 1,000,000,000,000 1012 giga G 1,000,000,000 109 mega M 1,000,000 106 kilo k 1,000 103 hecto h 100 102 deca da 10 101 BASE UNIT 1 100 deci d 0.1 10?1 centi c 0.01 10?2 milli m 0.001 10?3 micro μ 0.000001 10?6 nano n 0.000000001 10?9 pico p 0.000000000001 10?12 Metric ConversionsVolumeLMassgTimesEnergyJTemperatureK (no metric conversions; not a base 10 scale)10-1210-910-610-310-210-11001011021031061012pn?mcdBASEdahkMTpiconanomicromillicentidecidecahectakilomegatera+ ← → -Scientific NotationChemistry deals with very large and very small numbers. Consider this calculation: 0.000000000000000000000000000000663 x 30,000,000,000) ÷ 0.00000009116. Hopefully, you can see how awkward it is. Try keeping track of all those zeros! In scientific notation, this problem is: (6.63 x 10-31 x 3.0 x 1010) ÷ 9.116 x 10-8 = 2.18 x 10-13, AND it is much easier to solve!It is now much more compact, it better represents significant figures, and it is easier to manipulate mathematically. The trade-off, of course, is that you have to be able to read scientific notation. This lesson shows you (1) how to write numbers in scientific notation and (2) how to convert to and from scientific notation. As you work, keep in mind that a number like 9.116 x 10-8 is ONE number (0.00000009116) represented as a number 9.116 and an exponent (10-8). Format for Scientific NotationUse to represent positive numbers only. Every positive number X is written as: 1 < N < 10) x 10some positive or negative integer, where N represents the numerals of X with the decimal point after the first nonzero digit. A decimal point is in standard position if it is behind the first non-zero digit. Let X be any number and let N be that number with the decimal point moved to standard position. Then: If 0 < X < 1 then X = N x 10negative number If 1 < X < 10 then X = N x 100 If X > 10 then X = N x 10positive number Some examples of number three: 0.00087 becomes 8.7 x 10-4 9.8 becomes 9.8 x 100 (the 100 is seldom written) 23,000,000 becomes 2.3 x 107 0.000000809 becomes 8.09 x 10-74.56 becomes 4.56 x 100 250,000,000,000 becomes 2.50 x 1011 Note that standard position for the decimal place is always just to the right of the first non-zero digit in the number. Example 1: Convert 29,190,000,000 to scientific notation. Step 1 - Start at the decimal point of the original number and count the number of decimal places you move, stopping to the right of the first non-zero digit. Remember that's the first non-zero digit counting from the left. Step 2 - The number of places you moved (10 in this example) will be the exponent. If you moved to the left, it's a positive value. If you moved to the right, it's negative. The answer is 2.919 x 1010.Example 2: Write 0.00000000459 in scientific notation. Step 1 - Write all the significant digits down with the decimal point just to the right of the first significant digit. Like this: 4.59. Please be aware that this process should ALWAYS result in a value between 1 and 10. Step 2 - Now count how many decimal places you would move from 4.59 to recover the original number of 0.00000000459. The answer in this case would be 9 places to the LEFT. That is the number 0.000000001. Be aware that this number in exponential notation is 10-9. Step 3 - Write 4.59 times the other number, BUT, write the other number as a power of 10. The number of decimal places you counted gives the power of ten. In this example, that power would be 9. The correct answer to this step is: 4.59 x 10-9The answer is 4.59 x 10-9.Practice ProblemsConvert to scientific notation: 28,000,000 305,000 0.000000463 0.000201 3,010,000 0.000000000000057 20,100 0.00025 65,000,000,000,000,000 0.00854 x 1012 2101 x 10-16 305.1 x 107 0.0000594 x 10-16 0.00000827 x 1019 386 x 10-222511 x 1012 0.000482 x 10-12 0.0000321 x 1012 288 x 105 1234 x 1011 Molar Mass The molar mass of a substance is the weight in atomic mass units of all the atoms in a given formula. An atomic mass unit (amu) is defined as 1/12 the weight of the carbon-12 isotope. Carbon-12 is defined as weighing exactly 12 amu, and is the starting point for how much an atom weighs. For example, if you weigh 1/2 as much as C-12, you weigh 6. If you weigh twice as much, you weigh 24. The molar mass of a substance is needed to tell us how many grams are in one mole of that substance. The mole is the amount of substance of a system that contains as many elementary entities as there are atoms in 0.012 kilogram of carbon-12. On the periodic table, the amu (or g/mol) of an element is considered to be the same as the average atomic mass. This is known as the molar mass of the element, as it is the ‘mass of one mole of atoms of that element’ and/or the molecular mass. Point #1 - You need to know how many atoms of each element are in a substance in order to calculate its molar mass. For example H2O has two atoms of hydrogen and one atom of oxygen. H2O2 has two atoms each of oxygen and hydrogen. Mg(OH)2 has one atom of magnesiun and two each of oxygen and hydrogen. If a subscript follows an atom with no parenthesis, that number tells you how atoms are present. If parentheses are involved, you must multiply each subscript inside by the one that is outside. How many atoms of each element are in the following examples? KCl K1Cl1Fe2O3 Fe2O3Al(NO3)3 Al1N1 x 3 =3O3 x 3= 9NH4NO3 N2H4O3Al2(SO4)3 Al2S1 x 3 = 3O2 x 3 = 6Point#2 - You need to know the molar mass of each element in order to calculate the Molar mass of the substance. The molar mass of each element is found by examining the periodic table. Just below are typical entries in the periodic table for hydrogen and oxygen. For hydrogen the molar mass is 1.0079 and for oxygen it is 15.9994. How to Calculate the Molar Mass of a SubstanceStep 1: List the elements present in the compoundStep 2: Count the number of atoms of each element in the compound.Step 3: Determine the molar mass of each element by using the data from the periodic table. Step 4: Multiply the number of atoms of each element by the molar mass of that element.Step 5: Add the answers.Example: Al2(SO4)3 Step 1: List the elements present in the compound AlSOStep 2: Count the number of atoms of each element in the compound Al2S3 (1 x 3)O6 (2 x 3)Step 3: Determine the molar mass of each element by using the data from the periodic table. Al226.98S3 32.06O6 16.00Step 4: Multiply the number of atoms of each element by the molar mass of that element.Al2 x 26.98 = 53.96S3 x 32.06 = 96.18O6 x 16.00 = 96.00Step 5: Add the answers.Al2 x 26.98 = 53.96S3 x 32.06 = 96.18O6 x 16.00 = + 96.00 246.14g/molExample: H2O2 Step 1: List the elements present in the compoundHOStep 2: Count the number of atoms of each element in the compound.H2O2Step 3: Determine the molar mass of each element by using the data from the periodic table. H21.0079O216.00Step 4: Multiply the number of atoms of each element by the molar mass of that element.H2 x 1.0079 = 2.0158O2 x 16.00 = 32.00Step 5: Add the answers.H2 x 1.0079 = 2.0158O2 x 16.00 = + 32.0034.0158g/molSpecial Note about HydratesSuppose you are asked to calculate the molar mass of CuSO4 ? 5H2O. That dot DOES NOT mean multiply. You could approach this two ways: Add the atomic weights of one copper, one sulfur, nine oxygens, and ten hydrogens. Add the atomic weights of one copper, one sulfur, and four oxygens. Then add the molar mass of five H2O molecules. The answer is 249.68 amu. Practice ProblemsOn a separate sheet of paper calculate the Molar mass for the following compounds:AlCl3 Ba(SCN)2 LiHBa(BrO3)2AlBr3HClTeF4K2SHg2Cl2P2O5K2SO4PbS NH4ClSnI4Cr2(SO3)3NH4NO3NaClCu2OKH2PO4KOHAl(MnO4)3Ba(OH)2LiIAgI C2H5NBrMole Conversions: Given Grams, Convert to MolesIn chemistry, the mole is the standard measurement of amount. However, balances DO NOT give readings in moles. Balances give readings in grams. So the problem is that, while we compare amounts of one substance to another using moles, we must also use grams, since this is the information we get from balances. There are three steps to converting grams of a substance to moles. Step 1: Determine how many grams are given in the problem. Step 2: Calculate the molar mass of the substance. Step 3: Divide step one by step two. Example: Convert 25.0 grams of KMnO4 to moles. Step 1: The problem will tell you how many grams are present. Look for the unit of grams. The problem gives us 25.0 grams. Step 2: Calculate the molar mass of the substance. The molar mass of KMnO4 is 158.034 grams/mole. Step 3: You divide the grams given by the substance's molar mass: 21336001016000The answer of 0.158 mole has been rounded to three significant figures because the 25.0 value had the least number of significant figures in the problem. Example: Calculate the number of moles in 17.0 grams of H2O2 .17.0 grams are given in the text of the problem. The molar mass is 34.0146 grams/mole. 17g H202 x 1 mol H2O234.0146g H2O2 = 0.5 mols H2O2Practice Problems On a separate sheet of paper calculate the moles present in: 2.00 grams of H2O 75.57 grams of KBr 100. grams of KClO4 8.76 grams of NaOH 0.750 grams of Na2CO3 26.0 gram Ca(ClO4)25.08 gram XeF410.0 gram KAl(SO4)232.0 gram O210.0 gram V2O2.50 gram CoSO4 . 6H2O34.2 gram NH32.50 gram K2Cr2O724.0 gram CO9.00 gram H2SO410.00 gram Na2CO33.45 gram ZnCl259.3 gram SnF23.091 gram K2SO436.0 gram Na2CrO2 . 4H2O0.00500 gram XeO320.00 gram KOH15.0 gram PbO10.0 gram SO30.0089 gram IF750.00 gram KBr1.0 gram CO232.58 gram CuS1.0 x 102 gram KCl5.00 gram CaCO31.0 gram Ba(OH)212.25 gram Sr(HCO3)21.0 gram NaCl2.001 gram Al2O30.00860 gram Ca3(PO4)298.9 gram NaI2.00 x 10-3gram NH4NO314.0 gram N20.0010 gram Al(MnO4)Mole Conversions: Given Moles, Convert to GramsIn chemistry, the mole is the standard measurement of amount. When substances react, they do so in simple ratios of moles. However, balances give readings in grams. Balances DO NOT give readings in moles. So the problem is that, when we compare amounts of one substance to another using moles, we must convert from grams, since this is the information we get from balances. There are three steps to converting moles of a substance to grams: Step 1: Determine how many moles are given in the problem. Step 2: Calculate the molar mass of the substance. Step 3: Multiply step one by step two. Example: Convert 2.50 moles of KClO3 to grams. Step 1: 2.50 moles is given in the problem. Step 2: The molar mass for KClO3 is 122.550 grams/mole. Step 3: 2.50 moles x 122.550 grams/mole = 306.375 gramsPractice ProblemOn a separate sheet of paper calculate the grams present in: 0.100 moles of KI 1.500 moles of KClO 0.750 moles of NaOH 3.40 x 10-5 moles of Na2CO3 2.55 mole Cu2CrO41.95 mole HNO32.00 mole HC2H3O210.0 mole NaCl2.20 mole SnCl25.00 mole Ag2O3.00 mole H23.27 mole O21.55 mole KrF20.100 mole NH30.00550 mole CH40.100 mole H2O0.500 mole CaCO30.300 mole H3PO41.500 mole K2SO40.0010 mole H2SO45.0 mole NH4OH4.50 mole Na2O0.30 mole HCl0.00200 mole Na2SO4Avogadro Number Calculations: How Many Atoms or Molecules?The value for Avogadro's Number is 6.022 x 1023 mol-1. It can be used to determine the number of molecules, formula units, or particles of a given sample of substance.The value for Avogadro's Number is 6.022 x 1023/ mol Mass to Atoms/ Molecules/ Particles/ Formula UnitsStep 1. Define the given mass.Step 2: Determine the molar mass of the known. Step 3: Divide the mass by the molar mass.Step 4: Multiply the answer in Step 3 by Avogadro’s Number.Example: Ibuprofen, C13H18O2, is the active ingredient in many nonprescription pain relievers. Its molar mass is 206.29 g/mol. If the tablets in a bottle contain a total of 33 g of ibuprofen, how many moles of ibuprofen are in the bottle? How many molecules of ibuprofen are in the bottle?Step 1. Define the given mass. 33 g of C13H18O2, Step 2: Determine the molar mass of the known. C12.011g/mol x 13 = 156.143g/mol H 1.007g/mol x 18 = 18.126g/mol+ O 15.999g/mol x 2 = 31.998g/mol 206.267g/molStep 3: Divide the given mass by the molar mass. 33g ibuprofen x 1 mol ibuprofen = 0.160 mols ibuprofen 206.267gStep 4. Multiply the answer in Step 3 by Avogadro's number. 0.160 mols ibuprofen x 6.022 x 1023 molecules = 9.634 x 1022 molecules ibuprofen1 mol Practice Problems Convert the following to the correct number of atoms/ molecules/ particles/ formula units62.0 g of barium to atoms7.62 x 1011 g of lanthanum to molecules0.697 g of neon to particles 0.000 000 020 g beryllium to formula units54.0 g of aluminum to molecules69.45 g of calcium to atoms0.697 g of selenium to particles20 g uranium to atoms909.7 g ammonium phosphate to molecules3621. g copper (II) nitrate to formula units2.95 x 102 g cobalt (IV) chloride to molecules69.45 g of nitrogen dioxide to atoms9.991 x 1031 g carbon tetraiodide to formula units3.00 g of boron tribromide to atoms0.472 g of sodium fluoride to molecules7.50 x 102 g of heptanol, C7H15OH, to particlesAtoms/ Molecules/ Particles/ Formula Units to MassStep 1. Define the given number of Atoms/Molecules/Particles.Step 2: Divide the given number of Atoms/Molecules/Particles by Avogadro’s number.Step 3: Determine the molar mass of the compound.Step 4. Multiply the answer in Step 2 by the molar mass.Example 2: What is the mass of 3.022 x 1023atoms of NaCl?Step 1. Define the given number of Atoms/Molecules/Particles/Formula Units 3.022 x 1011atoms NaClStep 2: Divide the given number of Atoms/Molecules/Particles by Avogadro’s number.3.022 x 1023atoms NaCl = 0.50 mols NaCl 6.022 x 1023atoms Step 3: Determine the molar mass of the compound.Na 1 x 22.99 = 22.99Cl 1 x 35.45 = 35.45 58.44g/molStep 4. Multiply the answer in Step 2 by the molar mass.0.50 mols NaCl x 58.44g NaCl = 29.22g NaClPractice Problems:Calculate the number of g in each of the following. 3.697 x 1025 atoms of uranium 3.01 x 1023 formula units of hydrogen 4.068 x 1022 molecules of mercury 6 920 000 000 000 000 particles of cadmium 6.022 x 1024 atoms of tantalum 3.01 x1021 formula units of cobalt 1.506 x 1024 molecules of argon 1.20 x 1025 particles of helium 1.77 x 1023 atoms of sodium sulfide, Na2S 8.96 x 1020 formula units of nickel (II) nitrate, Ni(NO3)23.0 x 1021 molecules of acrylonitrile, CH2CHCN 8.22 x 1023 particles of oxygen tribromide, OBr34.05 x 1029 formula units of aluminum sulfate, Al2(SO4)32.94 x 1017 atoms of nitrogen trioxide, NO3 0.000 025 molecules of uranium (VI) chloride, UCl6 88 300 000 000 000 particles of iron (III) oxide, Fe2O3 ................
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