6 Mole Concept. g mol. g mol. g mol ) + 1( g : mol ratios are the units of molar mass. It does not matter which unit is on the

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1 What is a e? 6 Mole Concept The nature of chemistry is to chane one ecule into one or more new ecules in order to create new substances such as paints, fertilizers, food additives, medicines, etc. When a chemist is performin a reaction to make the new product, the chemist must know how much of the startin reactants to place in the reaction vessel. You know from your experience in the lab, that you can determine how much of a substance you have by determinin the mass or volume of the substance. However, atoms and ecules react not by mass or volume, but instead by number. This is exemplified when we write coefficients to balance chemical equations or subscripts in a chemical formula. These numbers are counts and have the unit of e. A e is a unit of countin similar to a dozen, except where a dozen is 1 of anythin a e is 6.0 x 10 3 of anythin. This number is known as Avoadro s number and is named after Amadeo Avoadro whose work in atomic ecular theory of ases led to the discovery of the number. Avoadro's number allows the chemist to determine the number of atoms or ecules in a sample. However, the measured size of a sample is still typically determined by mass on a balance. It is up to the chemist to interconvert between mass and es the count. This conversion is done with a conversion factor known as ar mass, M, which is the mass of 1 and has the units of /. Molar mass is determined by addin the atomic masses for each atom, not type, but each atom. For example, the ar mass of water, H O, is determined by: Conversions H atoms + 1 O atom = = 18.0 Chemists use a variety of ratios such as ar mass to interconvert between various quantities such as mass with the number of es. These ratios are known as conversion factors. Generally speakin a conversion between two units is done by multiplyin the iven unit by the conversion factor, which equals the wanted unit. wanted unit #.# iven unit wanted unit eq 1 iven unit Conversion equations are not sentences. That is, they are never written left to riht, but instead are written from the outside to the inside. One way to think about it, is that you need know where you are the iven, then where you want to o the wanted and finally you can decide how to et there the conversion factor. Overall, you first write the iven number #.# and the iven unit on the left, followed by an empty set of -- s with the fraction bar in it. Second write the wanted unit on the riht. Finally, fill in the s with the units: iven unit on the bottom so it cancels, wanted unit on the top so you have an equality. No numbers are in the s at this time. Exploration of the unit ratio of the conversion factor will determine the numbers that are to be substituted. Realize that it is not unusual to have difficulty in focusin on the units rather than on the numbers. Once master this technique of dimensional that is inherent in unit conversions is very useful and can be applied virtually anywhere in science. For e conversions there are really only three potential ratios of units that you will use for conversion factors. 1. or : ratios are the units of ar mass. It does not matter which unit is on the top or the bottom, the ratio is still ar mass. You calculate ar mass from the periodic table, by summin the atomic masses of all the atoms in the ecule. Molar mass is defined as the number of rams that are in one e of a substance, so the numerical answer that you calculate always oes with the rams and a 1 always oes with the es. The earlier example showed that the ar mass of ar mass of H O is 18.0/. In a conversion factor this is written: A B. or : ratios relate the number of one type of ecule that are reactin with B A another type of ecule. This relationship is expressed by the coefficients in a balanced chemical equation. Thus, the values that are substituted into the conversion factor are the coefficients for each ecule, which are found in the balanced chemical equation. For instance, consider NaN3 Na + 3N. The e ratio between NaN 3 and N would be: 18.0 HO 1 H O or NaN 3 N 3 1 HO 18.0 H O or 3 N NaN 3

2 ecules or ecules 6 Mole Concept The ratio of ecules to es is by definition 6.0 x 10 3 ecules : 1 e. Stoichiometry Multi-Step Conversion Problems ecules 1 This ratio is written or Stoichiometry is the relationship between two species based on the e ratio between them in a balanced chemical equation. Often in hih school, stoichiometry refers to problems that relate the mass of one compound to the mass of another compound, but technically any unit, especially volume for ases, can be used so lon as the e ratio is used. These conversions become more difficult because no one conversion factor can be used to solve the problem. When the iven and wanted units are assined, there is no one relationship between them so one conversion factor cannot be used. The trick is to separate the iven and wanted units throuh multiple parentheses,. The iven unit is still on the bottom of the fraction, but this time in the first parentheses and the wanted unit is still on top, but in the last parentheses. Additional units are added to express the relationships of ar mass, e-toe ratio, etc. This is a very rouh outline, a few examples will elucidate the pattern. Convert rams A rams C 3-step conversions, mass mass Molecules react by count es and not by mass. Therefore the followin sequence must be done to convert rams of one substance to rams of another substance. # A A C C This is a sequence of three conversions: A A, then A C, then C C. The conversions are the same as the one-step conversions studied previously, there are just three of them in a row. They may be written as three problems, but are more commonly and easily expressed as three conversion factors as follows. iven # 1 A # C #.## C A #.## A # A 1 B The units tell us that the first and third conversion factors are ar masses, so the values are calculated from the periodic table. The second conversion factor is a : ratio, so here the coefficients will be used for the values. Notice that the units all cancel on the diaonal which will serve as a quick way to check your work; if the units don t line up and cancel on the diaonal, then one of your conversion factors is probably upside-down. You can use this scheme to convert rams rams of any substances in a balanced chemical equation. not just rams of a reactant to rams of a product. It should be pointed out that whenever you need to relate an amount of one substance in a chemical equation with an amount of another substance in that equation; you are really doin a e-to-e conversion. You just have to do an extra step at the beinnin to et to es and then an extra step at the end to et away from the es. Convert rams A C or A rams C -step conversions Conversions between rams of one substance and es of another substance are -step conversion problems: one step to interconvert rams and es and one for the e to e. These problems are really short versions of the solution outlined in eq. These -step conversions utilize the same unit conversion scheme mentioned previously: A A C C. Generally speakin identify where the iven unit is in the series, and then identify where the wanted unit is in the series, and do those conversions. For example, rams A C scheme A A C C iven A wanted C Overall conversions 's: A A C So, do the first TWO 's of eq above. A rams C scheme A A C C iven A wanted C Overall conversions 's: A C C So, do the last TWO 's of eq above ecules wanted C eq

3 6 Mole Concept Example of Multi-Step Conversion Problems ex-1 If 16.0 of aluminum oxide, Al O 3, decomposes, then how many rams of oxyen, O, will be formed? ex- Al O 3 4 Al + 3 O 16.0 Al O 1 Al O 3 O 00 O 7.54 O AlO3 AlO3 1 O If 0.50 of aluminum oxide, Al O 3, decomposes, then how many rams of aluminum, Al, will be formed? Al O 3 4 Al + 3 O ex-3 How many es of aluminum oxide, Al O 3, must react to form 7.50 of oxyen, O? Limitin Reactant 4 Al 6.98 Al 0.50 Al O 15 Al In a chemical reaction involvin more than one reactant often the two reactants are present in their e ratios from the balanced chemical equations stoichiometric ratio. When the reactants are not in their stoichiometric ratio, then the limitin reactant is the reactant that runs out first, limitin the amount of products you can make. This is not necessarily the reactant present in the fewest es, but is instead the reactant that is present in the smallest amount based on the stoichiometric ratio. The calculation is straiht forward; for each reactant divide the number of es by the coefficient. The reactant with the smallest number is the limitin reactant. It is the limitin reactant that is then used to calculate the amount of product produced via the previous rules of stoichiometry. Focusin on the units, one can see in the examples that to determine the number of es of the product simply multiply by the product coefficient example 1 or to determine the mass of product, multiply by the product coefficient and ar mass example. Example of Limitin Reactant Problems 3 AlO3 1 Al 7.50 O 1 O Al O Al O O 3 O ex-1 Determine the limitin reactant when 0.05 e of nitroen, N, reacts with e of hydroen, H, accordin to the equation: N + 3H NH 3 N : H : 0.03 smallest, L. R Determine the es of ammonia, NH 3, produced when 0.05 N, reacts with H NH NH Because H is the limitin reactant 3 3 If the number of es of each reactant is not known, then if must be found from the iven information. Consider the followin example in which the es are not iven, but instead the mass of each reactant is iven. To solve, the mass in rams must first be converted to es, prior to dividin by the coefficient. ex- Determine the limitin reactant when.0 A reacts with.0 S 8: 16A + S 8 8A S..00 A 1 A A.00 S8 1 S S8 8 Determine the mass of A S produced when.0 A reacts with.0 S 8. Because A is the limitin reactant A : smallest, L. R 16 A S8 : S AS A S 1 A S A S

4 6 Mole Concept How to Determine Percent Composition by Mass from a Chemical Formula To calculate the percent by mass of one type of atom in a compound you calculate the percentae of that atom s ar mass in the overall ar mass of the compound. This will be a fraction with the denominator bein the ar mass of the compound, while the numerator is the term of the atom in the ar mass calculation. This is best demonstrated in the followin example. Example of Percent Composition Problems Calculate the percent of chlorine by mass in calcium chloride, CaCl. How to Determine an Empirical Formula from Percent Composition Data 1 st Percentaes become rams. This is because we can assume a 100 sample because the formula will be the same no matter what the sample size. {you don t have to show this step} nd Convert rams to es by dividin by the rams by the atomic mass from the periodic table. 3 rd Develop e ratios by dividin each number of es by the smallest. Note: if the decimal is you must multiply the ratio by a constant to et a whole number, then multiply all ratios by this constant. 4 th Write the formula by usin the whole numbers as subscripts. Example of Empirical Formula Problems Determine the empirical formula of a compound that is 40.0 % C, 6.7% H and 53% O. 1 st C: 40.0 % 40.0 H: 6.7 % 6.7 O: 53 % 53 nd 3 rd 4 th C: C 1H O 1 H: O: Cl % Cl x 100% 1Ca Cl % Cl x 100% % Cl 689% How to Determine the Molecular Formula from the Empirical Formula 1 st Divide the ar mass of the ecular formula iven in problem by the ar mass of the empirical formula calculated from the empirical formuls and round to a constant, nd Multiply the empirical formula subscripts by this constant. Example of Determinin the Molecular Formula What is the ecular formula of a substance with an empirical formula of C1HO1 and a ecular ar mass of 180 /? Molecular Molar Mass Empirical Molar Mass {C 1H O 1} x 6 = C 6H 1O 6

5 6 Mole Concept SUMMARY: Mole Concept Terms, Calculations and Examples Conversions: conversions conversions conversions conversions divide the iven by the ar mass sum the atomic masses in the periodic table example: convert.00 of H O to es of H O multiply the iven by the ar mass sum the atomic masses in the periodic table example: convert of CH 4 to of CH 4 multiply the iven by the e/e ratio coefficients in the chemical equation example: NaCl + Br NaBr + Cl How many es of bromine react with.5 es of sodium chlorine? multiply the iven by the three parentheses of a stoichiometry problem example: NaCl + Br NaBr + Cl How many rams of chlorine can be made from.5 of sodium chloride? ecule conversions multiply the iven by Avoadro s number example: How many ecules are in 0.01 of H O? ecule conversions divide the iven by Avoadro s number example: How many es are in 4.05x 10 1 ecules of CH 4? Empirical Formula Excess Reactant Limitin Reactant Molarity, M Br 1.3Br.5 NaCl.5 NaCl NaCl 1 NaCl 1Cl 70.90Cl 1.5 Cl NaCl NaCl 1Cl x10 ecules 6.0 x10 3 ecules 1 1 formulas written with the simplest ratio of atoms, not the exact ratio. example: ecular formula C 6H 1O 6 M = 180 / empirical formula C 1H O 1 M = 30 / the reactant that will have an amount remainin when the reaction is complete the reactant that will be completely consumed by the reaction x10 ecule the concentration of a solute via es of solute per total volume of solution. x ecule Percent Composition Units are L ; and arity is symbolized by [ ] s around a formula, e.. [NaF] the percentae by mass of one element in a compound example: What is the percent by mass of carbon, C, in fructose, C 6H 1O 6? 6C C 1H 6O % C x100% % C x100% % C 6.66% Stoichiometry Yield Experimental Yield Theoretical Yield Percent Yield any relationship between species in a balanced chemical equation that utilizes the e ratio of those species in the equation the amount of product made from a iven amount of reactant the amount of product actually isolated in the experiment the amount of product calculated from a iven amount of reactant see stoich. the ratio of the actual amount of product isolated to the theoretical yield of product; expressed as a percentae: actual amount 100% % yield theoretical yield

6 Math and Measurement 6 Mole Concept The numbers used in chemistry represent actual values of measurements. Like all measurements, these numbers have error associated with them the error may be small or lare, but is still error. The error exists because the last place in any measurement is an estimate/uess. As an example, we will measure the lenth of one side of a rectanle usin the scale in the diaram below The lenth is more than 3 but less than 4. A ood estimate would be or The final place in the number is not a wild uess, but it is a uess none the less. So there is error in the final place, and this error ets carried alon in any calculation that is done with that number. This can et confusin when doin math calculations with multiple numbers each with their own error. What kind of error ets carried over into the final answer? The worst error ets carried over the number that is least well known least number of sinificant fiures. Another example of error can be found in election polls. In such polls they state the percentae of the vote for each candidate and include the error. So picture an election with two candidates: Luke and Lexi. The polls list Luke at 4% and Lexi at 38% with an error of ±3%. This means that Luke s actual rane of the vote is anywhere from 45% - 39% 4+3 to 4-3, while Lexi s actual rane of the vote is anywhere from 41% - 35% 38+3 to 38- Therefore the pollin simultaneously suests a landslide victory for Luke {Luke 45% to Lexi 35%} to an upset for Lexi {Luke 39% to Lexi 41%}. The error clearly makes an impact on the value of the numbers. Measurin The final diit in a measured value is always an estimate/uess. For instance, if a thermometer is marked in increments of 1 C, then we can measure to the 0.1 C. Precision Reproducibility of the measurement. Also the exactness extent to which a measurement is known. e.. ±0.1 or ±0.01 etc. See sinificant fiures Accuracy Closeness of a measurement to the actual value Sinificant Fiures Diits in the number that represent the error in the measurement. This shows the precision of the value. e si. fis. is more precise than.4 si. fis. A diit is sinificant if it is a countin number, a zero between two sinificant fiures or a zero after a decimal and a sinificant fiure. A constant has an infinite number of sinificant fiures. When multiplyin: the answer has the least number of sinificant fiures When addin: the answer has the least number of decimal places Examples Consider the numbers: 60 and has 4 si fi 4.0 has 3 si fi Consider the numbers: 360 and has 3 si fi has 3 si fi Multiplyin: 17.0 x = = sf s x sf s so the answer must be rounded to si fi Multiplyin: 408 x 5.10 = =.0 x 10 4 sf s x 3 sf s so the answer must be rounded to 3 si fi to write to 3 si fi s, must use scientific notation Addin: = 53 rounded to 1 decimal place Percent Error Percentae by which a measurement differs from the actual value accepted value measured value % error accepted value 100%