1 T-6 Tutorial 2 FORMULAS, PERCENTAGE COMPOSITION, AND THE MOLE FORMULAS: A chemical formula shows the elemental composition of a substance: the chemical symbols show what elements are present and the numerical subscripts show how many atoms of each element there are in a formula unit. Examples: NaCl: one sodium atom, one chlorine atom in a formula unit CaCl 2 : one calcium atom, two chlorine atoms in a formula unit Mg 3 N 2 : three magnesium atoms, two nitrogen atoms in a formula unit The presence of a metal in a chemical formula indicates an ionic compound, which is composed of positive ions (cations) and negative ions (anions). A formula with only nonmetals indicates a molecular compound (unless it is an ammonium, NH 4 +, compound). Only ionic compounds are considered in this Tutorial. There are tables of common ions in your lecture text, p 56 (cations) and p 57 (anions). A combined table of these same ions can be found on the inside back cover of the lecture text. A similar list is on the next page; all formulas needed in this and subsequent Tutorial problems can be written with ions from this list. Writing formulas for ionic compounds is very straightforward: TOTAL POSITIVE CHARGES MUST BE THE SAME AS TOTAL NEGATIVE CHARGES. The formula must be neutral. The positive ion is written first in the formula and the name of the compound is the two ion names. EXAMPLE: Write the formula for potassium chloride. The name tells you there are potassium, K +, and chloride, Cl, ions. Each potassium ion is +1 and each chloride ion is -1: one of each is needed, and the formula for potassium chloride is KCl. "1" is never written as a subscript. EXAMPLE: Write the formula for magnesium hydroxide. This contains magnesium, Mg 2+, and hydroxide, OH, ions. Each magnesium ion is +2 and each hydroxide ion is -1: two -1 ions are needed for one +2 ion, and the formula for magnesium hydroxide is Mg(OH) 2. The (OH) 2 indicates there are two OH ions. In a formula unit of Mg(OH) 2, there are one magnesium ion and two hydroxide ions; or one magnesium, two oxygen, and two hydrogen atoms. The subscript multiplies everything in ( ). EXAMPLE: Write the formula for aluminum sulfate. This contains aluminum, Al 3+, and sulfate, SO 4 2, ions. The lowest common multiple of 3 and 2 is 6, so we will need six positive and six negative charges: two Al 3+ and three SO 4 2 ions, and the formula for aluminum sulfate is Al 2 (SO 4 ) 3. Then, in a formula unit of Al 2 (SO 4 ) 3 there are two aluminum ions and three sulfate ions; or two aluminum, three sulfur, and twelve oxygen atoms.
2 COMMON POSITIVE IONS COMMON NEGATIVE IONS COMMON NEGATIVE IONS H + hydrogen [Fe(CN) 6 ] 3 ferricyanide C 2 H 3 O 2 acetate NH 4 + ammonium [Fe(CN) 6 ] 4 ferrocyanide CN cyanide Li + lithium 3 PO 4 phosphate CNO cyanate Na + sodium 2 HPO 4 hydrogen phosphate SCN thiocyanate K + potassium H 2 PO 4 dihydrogen phosphate ClO hypochlorite Mg 2+ magnesium 2 CO 3 carbonate ClO 3 chlorate Ca 2+ calcium HCO 3 hydrogen carbonate ClO 4 perchlorate Sr 2+ strontium 2 SO 3 sulfite IO 3 iodate Ba 2+ barium HSO 3 hydrogen sulfite MnO 4 permanganate Al 3+ aluminum 2 SO 4 sulfate NO 2 nitrite Sn 2+ tin(ii) HSO 4 hydrogen sulfate NO 3 nitrate Sn 4+ tin(iv) 2 S 2 O 3 thiosulfate OH hydroxide Pb 2+ lead(ii) 1 2 CrO 4 chromate IO 4 periodate Bi 3+ bismuth 2 Cr 2 O 7 dichromate H hydride Cr 3+ chromium(iii) 2 O 2 oxide F fluoride Mn 2+ manganese(ii) 3 2 O 2 peroxide Cl chloride Fe 2+ iron(ii) S 2 sulfide Br bromide Fe 3+ iron(iii) HS hydrogen sulfide I iodide Co 2+ cobalt(ii) 4 Ni 2+ nickel(ii) 5 Cu + copper(i) Cu 2+ copper(ii) 1 There is also a lead(iv) Ag + silver 2 There is also a chromium(ii) Zn 2+ Cd Hg 2 2+ Hg 2 zinc 3 There is also a manganese(iii) cadmium 4 There is also a cobalt(iii) mercury(i) 5 There is also a nickel(iii) mercury(ii)
3 T-8 PERCENTAGE COMPOSITION: Imagine a class of 40 boys and 60 girls: 100 students total. 40/100 of the students are boys and 60/100 girls; or, 40% boys and 60% girls. The "%" sign means percent, or parts per 100. Suppose a class has 10 boys and 15 girls, for a total of 25 students. If you want to find out how many boys there would be per 100 students, keeping the same ratio of boys to girls, the following proportion can be set up: 10 boys X boys = 25 students 100 students This means: "if there are 10 boys per 25 students, then there would be X boys per 100 students." Solving this equation, 10 boys X = x 100 students = 40 boys 25 students There would be 40 boys per 100 students, or 40% boys. Percent calculation is based on this kind of proportion. However, you do not have to set up the proportion each time you want to find percent composition. Simply divide the number of parts you are interested in (number of boys) by the total number of parts (number of students), and multiply by 100. This is exactly what we did to find X in the above example. DIVIDE THE PART BY THE WHOLE AND MULTIPLY BY 100 Suppose a mixture contains 12.4 g salt, 15.3 g sugar and 50.3 g sand, for a total 78.0 g of mixture. The percentage of each component in the mixture can be found by dividing the amount of each (PART) by the total amount (WHOLE) and multiplying by 100: 12.4 g 15.3 g x 100 = 15.9% salt x 100 = 19.6% sugar 78.0 g 78.0 g 50.3 g x 100 = 64.5% 78.0 g The sum of these percentages is 100%. Everything must total 100%.
4 T-9 Percentage is written without units but the units are understood to be "parts of whatever per 100 total parts." The "parts" and "whole" may be in any units as long as they are both in the same units. In the case of the above mixture, units could be put on the numbers and the percentages then used as conversion factors. For example, the 15.9% salt could become 15.9 g salt pounds mixture 15.9 tons salt or or g mixture 15.9 pounds salt tons mixture Et cetera. We do not usually care about composition of salt/sugar/sand mixtures, but we do care about percentage composition of pure chemical substances. Unless otherwise stated we always mean percent by mass (weight). The formula mass of a substance is the sum of the atomic masses of all the atoms in the formula. For example, the formula mass of sodium chloride, NaCl, 58.44, is the sum of the atomic mass of sodium, 22.99, and the atomic mass of chlorine, The formula mass of sodium sulfate, Na 2 SO 4, is found as follows: 2 Na: 2 x = g Na or lbs Na 1 S: 1 x = g S or lbs S 4 O: 4 x = g O or lbs O Formula mass = g Na 2 SO lbs Na 2 SO 4 Atomic mass and formula mass are in atomic mass units, but are usually written without units. However, since these are relative masses of the atoms any mass unit can be used, as shown by the two columns to the right of the arrows, above. In terms of atomic masses: in a total of parts by mass of sodium sulfate, parts are sodium, parts are sulfur, and parts are oxygen. In terms of a gram mass unit (molar mass): in a total of g of sodium sulfate, g are sodium, g are sulfur, and g are oxygen. In terms of pound mass unit: in a total of lbs of sodium sulfate, lbs are sodium, lbs are sulfur, and lbs are oxygen.
5 T-10 Percent composition can be determined with the mass ratios from the formula mass: Sodium: x 100 = 32.37% Na g Na or lbs Na Sulfur: x 100 = 22.57% S g S or lbs S Oxygen: x 100 = 45.06% O g O or lbs O Total = % g total lbs total As mentioned before, percentage is written without any units but the units are understood to be "parts of whatever per 100 total parts". Since these are relative masses any mass unit can be used, as shown by the two columns to the right of the arrows, above. In terms of percentage: in a total of parts by mass of sodium sulfate, parts are sodium, parts are sulfur, and parts are oxygen. In terms of a gram mass unit (molar mass): in a total of g of sodium sulfate, g are sodium, g are sulfur, and g are oxygen. In terms of a pound mass unit: in a total of lbs of sodium sulfate, lbs are sodium, lbs are sulfur, and lbs are oxygen. In the preceding calculations, the percent of each element in the substance was found. The percent of groups of elements may also be calculated. Suppose you want percent sulfate, SO 4 2, in sodium sulfate. The formula mass of sodium sulfate is and of that total = is sulfate. Thus, Sulfate: x 100 = 67.63% SO Notice: this is the sum of the percents sulfur and oxygen: 22.57% S % 0 = 67.63% SO 4 2. We may say sodium sulfate is 32.37% sodium, 22.57% sulfur, and 45.06% oxygen; or, 32.37% sodium and 67.63% sulfate. The total is 100% in either case.
6 T-11 We now have two sets of numbers which show mass relationships among the elements in sodium sulfate: (1) the formula mass and (2) the percentage composition. From the formula mass on page T-9 various mass ratios, that is, conversion factors, may be obtained: g Na lbs Na 2 SO g S or or and others g Na 2 SO lbs S g Na From the percentages on page T-10 various mass ratios may also be obtained: g Na lbs Na 2 SO g S or or and others g Na 2 SO lbs S g Na We can now solve problems like the following: EXAMPLE: What mass of sulfur is contained in 17.2 g of sodium sulfate? Use formula mass ratios on page T-9 for the appropriate conversion factor: g S 17.2 g Na 2 SO 4 x = 3.88 g S g Na 2 SO 4 OR, use percentage composition ratios on page T-10: g S 17.2 g Na 2 SO 4 x = 3.88 g S g Na 2 SO 4
7 T-12 EXAMPLE: What mass of sodium is combined with 10.0 g of sulfur in sodium sulfate? You may use a conversion factor from the formula masses OR percentage composition: OR g Na 10.0 g S x = 14.3 g Na g S g Na 10.0 g S x = 14.3 g Na g S Obviously, you do not have to calculate percent composition every time you do a problem like this: use formula mass ratios; if the percentage composition is given, then use it. THE MOLE: The most important concept you encounter this semester in Chemistry is the mole. The mole may be defined: The mole is the amount (mass) of a substance which contains the same number of units (atoms, molecules, ions) as there are atoms in exactly 12 mass units of the carbon-12, 12 C, isotope. It does not matter what mass units are used as long as the same mass units are used for the substance and the carbon-12. This is a strict definition of the mole, but it is operationally useless. A mole of a monoatomic element is the atomic mass taken with mass units; a mole of anything else is the formula mass taken with mass units. The kind of mole you get depends on the mass units you use. For example: grams Na = 1 gram-mole Na has the same number of Na atoms as the number of atoms in 12 grams of 12 C milligrams Cu = 1 milligram-mole Cu has the same number of Cu atoms as the number of atoms in 12 milligrams of 12 C grams Cl 2 = 1 gram-mole Cl 2 has the same number of Cl 2 molecules as the number of atoms in 12 grams of 12 C lbs I 2 = 1 lb-mole I 2 has the same number of I 2 molecules as the number of atoms in 12 lbs of 12 C grams NaCl = 1 gram-mole NaCl has the same number of NaCl formula units as the number of atoms in 12 grams of 12 C.
8 T tons Na 2 SO 4 = 1 ton-mole Na 2 SO 4 has the same number of Na 2 SO 4 formula units as the number of atoms in 12 tons of 12 C g H 2 O = 1 gram-mole H 2 O has the same number of H 2 O molecules as the number of atoms in 12 g of 12 C kg PCl 3 = 1 kilogram-mole PCl 3 has the same number of PCl 3 molecules as the number of atoms in 12 kg of 12 C. Chemists generally use the gram-mole, written mole, abbreviated mol. This is the mole your textbook uses and is the only one for which Avogadro's Number is applicable. In one grammole of a substance there is Avogadro's Number of units: x of them. Unless further qualification is noted the term "mole" is understood to mean the gram-mole. Each of the equalities above becomes a conversion factor: And so on g Na 1 mol Na g Na = 1 mol Na becomes or 1 mol Na g Na EXAMPLE: How many moles is g of sodium metal? 1 mol Na g Na x = mol Na g Na EXAMPLE: What is the mass of mol of sodium sulfate? 142 g Na 2 SO mol Na 2 SO 4 x = 51.4 g Na 2 SO 4 1 mol Na 2 SO 4 The number of significant figures used in the atomic mass or formula mass should be at least as many as in the given (measured) data. EXAMPLE: How many atoms of copper are in a 125 mg sample of copper metal? When a problem asks "how many atoms or molecules" or "what is the mass of an atom or molecule," you must use Avogadro's Number. The units associated with that Number then depend on the problem. In this problem you are looking for atoms of copper so the units will be:
9 T x atoms Cu 1 mol Cu The number of atoms of copper is related to moles of copper; moles of copper is related to grams of copper; the mass of copper given in the problem is in milligrams. The conversion sequence will be: mg Cu g Cu mol Cu atoms Cu 1 g Cu 1 mol Cu x atoms Cu 125 mg Cu x x x = 1000 mg Cu 63.5 g Cu 1 mol Cu 1.19 x atoms Cu EXAMPLE: What is the mass of one water molecule? This time the units associated with Avogadro's Number are x H 2 O molecules 1 mol H 2 O We are converting from a number of molecules (1, an exact number) to mass: 1 mol H 2 O 18.0 g H 2 O 1 H 2 O molecule x x = x H 2 O molecules 1 mol H 2 O 2.99 x g H 2 O If four significant figures had been used for the formula mass of water, 18.02, the answer would have been x g H 2 O. In this case the precision of the answer depends on the conversion factors, since the given data is an exact number. Chemical formulas should be considered in terms of moles of atoms combined: the formula NaCl means there is one mole of sodium combined with one mole of chlorine; the formula H 2 O means there are two moles of hydrogen combined with one mole of oxygen to form one mole of water.
10 T-15 The formula mass for sodium sulfate was calculated on page T-9. Referring to those numbers, the formula Na 2 SO 4 means: two moles sodium (45.98 g), one mole sulfur (32.06 g), and four moles oxygen (64.00 g) combine to form one mole of sodium sulfate ( g). Let us use this concept and do the first problem on page T-11 a different way: EXAMPLE: What mass of sulfur is contained in 17.2 g of sodium sulfate? 1 mol Na 2 SO 4 1 mol S 32.1 g S 17.2 g Na 2 SO 4 x x x = 3.89 g S 142 g Na 2 SO 4 1 mol Na 2 SO 4 1 mol S Please notice: The same numbers are used in the same places, so the overall arithmetic is the same. The thought process is different: on page T-11 we thought in terms of mass ratios, now we are thinking in terms of mole ratios. This answer is slightly different from the one obtained on page T-11. Can you see why? EMPIRICAL (SIMPLEST) FORMULA: The empirical formula is the formula expressed as the lowest whole number ratio of atoms; it may or may not be the same as the formula. The empirical formula and formula of water, H 2 O, are the same; the empirical formula of phosphorous pentoxide is P 2 O 5, but the formula is P 4 O 10. The empirical formula is derived solely from experimental, percentage composition data. On page T-10 we calculated percentage composition from a formula; let us now calculate an empirical formula from percentage composition data. Fundamental to this calculation is the fact: chemical formulas show mole ratios of elements. EXAMPLE: Calculate the empirical formula for a compound with the analysis: 32.37% Na 22.57% S 45.06% O Remember percentage means "parts of whatever per 100 total parts," so let us translate these percentages into grams of each element out of a total 100 grams of compound. Then convert these masses to moles: 1 mol Na g Na x = mol Na g Na 1 mol S g S x = mol S g S 1 mol O g O x = mol O g O
11 T-16 Please note: these are the number of moles of each element contained in g of compound. Now find the ratio of moles of each element to the smallest number of moles, by dividing each number of moles by the smallest: mol Na = mol Na / mol S mol S mol S = mol S mol O = mol O / mol S mol S These ratios say there are 2 mol Na per 1 mol S; and 4 mol O per 1 mol S. The lowest wholenumber ratio of moles is: Na 2 S 1 O 4 ; and since we do not write "1" as a subscript, the formula is: Na 2 SO 4. Sometimes these two steps do not give whole number ratios. In that case an additional step is needed, as shown below. EXAMPLE: Calculate the empirical formula for a compound with the analysis: 29.1% Na 40.5% S 30.4% O Let us do as with the previous EXAMPLE: (1) translate the percents into grams of the element per 100 g of compound and find moles of each element; then (2) find the ratio of moles of each element to the smallest number of moles: 1 mol Na 1.27 mol Na 29.1 g Na x = 1.27 mol Na = 1.01 mol Na/mol S 23.0 g Na 1.26 mol S 1 mol S 1.26 mol S 40.5 g S x = 1.26 mol S = g S 1.26 mol S 1 mol O 1.90 mol O 30.4 g O x = 1.90 mol O = 1.51 mol O/mol S 16.0 g O 1.26 mol S
12 T-17 These ratios say there is 1 mol Na per 1 mol S, and 1.5 mol O per 1 mol S, which are not whole number ratios. To make them whole number ratios we must multiply each by 2, giving ratios of 2 mol Na per 2 mol S and 3 mol O per 2 mol S; and the formula is Na 2 S 2 O 3. Thus, in cases where steps (1) and (2) do not lead to whole numbers, (3) multiply each ratio by some integer (2, 3, etc.) to make all the ratios whole numbers. 1) Write formulas for the following: a) bismuth nitrate j) zinc phosphate s) copper(ii) oxide b) calcium hydride k) magnesium carbonate t) lithium nitrate c) mercury(ii) cyanide l) tin(ii) chloride u) aluminum sulfide d) strontium iodide m) lead(ii) bromide v) iron(ii) phosphate e) tin(iv) hydroxide n) ammonium dichromate w) barium sulfate f) potassium sulfite o) nickel(ii) chromate x) silver nitrate g) sodium carbonate p) cadmium cyanate y) copper(i) sulfide h) cobalt(ii) nitrite q) mercury(i) chloride z) iron(iii) sulfate i) chromium(iii) iodide r) manganese(ii) periodate 2) Calculate the percentage composition, by mass, of the following. Use four significant figures. a) potassium phosphate e) zinc chloride b) ammonium cyanide f) aluminum nitrate c) sodium hydrogen sulfate g) chromium(iii) hydroxide d) potassium perchlorate h) copper(ii) carbonate 3) Using the percentages from problem (2) calculate the following: a) grams of phosphate in 65.5 g of potassium phosphate b) milligrams of nitrogen in mg of ammonium cyanide c) pounds of sodium in 10.0 lbs of sodium hydrogen sulfate d) grams of chlorine in 37.5 g of potassium perchlorate e) tons of zinc in tons of zinc chloride f) tons of nitrate in 125 tons of aluminum nitrate g) milligrams of oxygen in g chromium(iii) hydroxide h) pounds carbon in 138 lbs copper(ii) carbonate
13 T-18 4) Convert each of the following to moles: a) g of iron(iii) sulfate b) 3.67 x 10 2 g of lead(ii) nitrate c) x 10 3 g of tin(ii) bromide d) kg of ammonium chromate e) 175 g of manganese(ii) carbonate f) 8.5 x 10 6 g of zinc oxide g) mg of barium sulfate h) 3.55 x 10 5 mg of aluminum chloride 5) Calculate the mass (in grams) of each of the following: a) 1.00 mol of barium sulfate b) mol of zinc nitrate c) mol of sodium hydrogen carbonate d) mol of potassium permanganate e) 1.67 x 10 2 mol of sodium hydroxide f) x 10 4 mol of copper(i) chloride g) mol of magnesium chloride h) 1.45 x 10 2 mol of calcium bromide 6) Calculate the following: a) grams of manganese in 1.55 g of manganese(ii) nitrate b) tons of phosphorous in 5.73 tons of potassium phosphate c) moles of sulfur in 7.33 g of aluminum sulfate d) pounds of iron in 12.6 lbs of iron(ii) phosphate e) moles of nitrogen in 5.55 g of lead(ii) nitrate f) grams of nitrogen in 3.95 g of ammonium phosphate g) grams of iron in 1.37 kg of iron(iii) sulfide h) moles of oxygen in g of iron(iii) sulfate 7) Calculate the following: a) number of sodium atoms in 6.79 g of sodium metal b) number of sodium ions in 7.77 g of sodium phosphate c) number of water molecules in 15.5 mg of water d) number of cobalt atoms in mg of cobalt metal e) number of ammonia molecules, NH 3, in 375 kg of ammonia f) number of gold atoms in 1.00 x 10 6 mg of gold g) number of ozone molecules, O 3, in 1.73 x 10 8 g of ozone h) mass, in grams, of ten water molecules i) mass, in grams, of one mercury atom j) mass, in milligrams, of three sulfuric acid, H 2 SO 4, molecules k) mass, in kilograms, of 1.25 x sulfate ions l) mass, in kilograms, of 2.57 x ammonium ions
14 T-19 8) Calculate the empirical formulas for the compounds with the following analyses: a) 34.4% Fe 65.6% Cl b) 29.4% Ca 23.6% S 47.0% O c) 28.7% K 1.5% H 22.8% P 47.0% O d) 28.0% Fe 24.1% S 48.0% O e) 88.8% Cu 11.2% O f) 72.3% Fe 27.7% O g) 69.9% Fe 30.1% O h) 11.1% N 3.2% H 41.3% Cr 44.4% O i) 68.4% Cr 31.6% O Answers to Problems 1) a) Bi(NO 3 ) 3 h) Co(NO 2 ) 2 o) NiCrO 4 u) Al 2 S 3 b) CaH 2 i) CrI 3 p) Cd(CNO) 2 v) Fe 3 (PO 4 ) 2 c) Hg(CN) 2 j) Zn 3 (PO 4 ) 2 q) Hg 2 Cl 2 w) BaSO 4 d) SrI 2 k) MgCO 3 r) Mn(IO 4 ) 2 x) AgNO 3 e) Sn(OH) 4 l) SnCl 2 s) CuO y) Cu 2 S f) K 2 SO 3 m) PbBr 2 t) LiNO 3 z) Fe 2 (SO 4 ) 3 g) Na 2 CO 3 n) (NH 4 ) 2 Cr 2 O 7 2) a) 55.26% K; 14.59% P; 30.15% O for K 3 PO 4 b) 63.59% N; 9.152% H; 27.26% C for NH 4 CN c) 19.15% Na; % H; 26.71% S; 53.30% O for NaHSO 4 d) 28.22% K; 25.59% Cl; 46.19% O for KCl0 4 e) 47.98% Zn; 52.02% Cl for ZnCl 2 f) 12.67% Al; 19.73% N; 67.60% O for Al(NO 3 ) 3 g) 50.47% Cr; 46.59% O; 2.935% H for Cr(OH) 3 h) 51.43% Cu; 9.720% C; 38.85% O for CuCO 3 3) a) g PO 4 e) tons Zn b) mg N f) tons NO 3 c) 1.92 lbs Na g) mg O d) 9.60 g Cl h) 13.4 lbs C
15 T-20 4) a) mol Fe 2 (SO 4 ) 3 e) 1.52 mol MnCO 3 b) 1.11 x 10 4 mol Pb(NO 3 ) 2 f) 1.0 x 10 7 mol ZnO c) mol SnBr 2 g) x 10 6 mol BaSO 4 d) mol (NH 4 ) 2 CrO 4 h) 2.66 mol AlCl 3 5) a) 233 g BaSO 4 e) g NaOH b) g Zn(NO 3 ) 2 f) g CuCl c) 13.0 g NaHCO 3 g) g MgCl 2 d) g KMnO 4 h) 2.90 x 10 4 g CaBr 2 6) a) g Mn in Mn(NO 3 ) 2 e) mol N in Pb(NO 3 ) 2 b) ton P in K 3 PO 4 f) 1.11 g N in (NH 4 ) 3 PO 4 c) mol S in Al 2 (SO 4 ) 3 g) 735 g Fe in Fe 2 S 3 d) 5.90 lbs Fe in Fe 3 (PO 4 ) 2 h) mol O in Fe 2 (SO 4 ) 3 7) a) 1.78 x Na atoms g) 2.17 x O 3 molecules b) 8.56 x Na 1+ ions in Na 3 PO 4 h) x g H 2 O c) 5.19 x H 2 O molecules i) x g Hg d) x Co atoms j) x mg H 2 SO 4 e) 1.33 x NH 3 molecules k) 1.99 x kg SO 4 f) 3.06 x Au atoms l) kg NH 4 1 8) a) FeCl 3 b) CaSO 4 c) KH 2 PO 4 d) Fe 2 (SO 4 ) 3 e) Cu 2 O f) Fe 3 O 4 g) Fe 2 O 3 h) (NH 4 ) 2 Cr 2 O 7 i) Cr 2 O 3
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Inorganic Nomenclature In order to efficiently be able to discuss chemicals and reactions, we need to understand a common system of chemical nomenclature. Nomenclature is crucial as you further your studies
Balancing Chemical Equations Aluminum reacts with copper(ii) chloride, CuCl 2, to form copper metal and aluminum chloride, AlCl 3 1. Identify the reactants and products. Aluminum and copper(ii) chloride
Example Exercise 8.1 Evidence for a Reaction Which of the following is experimental evidence for a chemical reaction? (a) Pouring vinegar on baking soda gives foamy bubbles. (b) Mixing two solutions produces
Balancing Chemical Equations Worksheet Student Instructions 1. Identify the reactants and products and write a word equation. 2. Write the correct chemical formula for each of the reactants and the products.
Activity One: Binary Ionic Compounds Composed of Main Group Elements This activity focuses on the nomenclature rules for ionic compounds composed of two different elements from select groups on the periodic
ANSWERS: Formula writing and nomenclature of inorganic compounds 1. Determine the oxidation number of S in each of the following compounds: a) Na 2 S 2 O 3 ans. a) +2 b) H 2 SO 3 b) +4 c) SO 2 c) +4 d)
Reaction Guidelines Students choose five of the eight reactions. Only the answers in the boxes are graded (unless clearly marked otherwise). Each correct answer earns 3 points, 1 point for reactants and
AP Equations Review Write the formulas to show the reactants and the products for any FIVE of the laboratory situations described below. Answers to more than five choices will not be graded. In all cases,
WRITING CHEMICAL EQUATIONS 2004, 2002, 1989 by David A. Katz. All rights reserved. Permission for classroom used provided original copyright is included. David A. Katz Chemist, Educator, Science Communicator,
Naming Compounds Tutorial and Worksheet Since we use different methods in naming binary covalent (molecular) compounds and ionic compounds, the first step in naming or writing the formula of a compound
5 REDOX REACTIONS AND REDOX EQUATIONS DEFINITIONS Most of the reactions and their equations considered so far have been based upon a rearrangement of ions between the components of a mixture. Another very
Skills Worksheet Problem Solving Percentage Composition Suppose you are working in an industrial laboratory. Your supervisor gives you a bottle containing a white crystalline compound and asks you to determine
Chapter 3 Ionic compounds 3.1 Ions Atoms have the ability to gain or loose electrons and become ions. Ion: an electrically charged atom or group of atoms. Cation: an atom that has lost one or more electrons
Chapter 5 Page 1 Chapter 5: Molecules and Compounds Compounds-a pure substance composed of more than one type of where the combine chemically infixed, definite proportions. Contrast a compoundto a mixturewhere
Ionic Compounds Ionic compounds are made up of electrically equivalent numbers of cations and anions. Na + + Cl Y NaCl Ba 2+ + 2Cl Y BaCl 2 Cr 3+ + 3Cl Y CrCl 3 2Al 3+ + 3O 2 Y Al 2 O 3 Simple ionic compounds
WRITE IN THE CORRECT FORMULA THEN BALANCE THE CHEMICAL EQUATIONS 1. Hydrogen Oxygen Water METALLIC OXIDE WATER BASE 2. Sodium oxide Water Sodium hydroxide 3. Calcium oxide Water Calcium hydroxide 4. Potassium
6 Nomenclature of Inorganic Compounds Seashells are formed from calcium carbonate (CaCO3) aka limestone. This compound is also used in many dietary supplements. Chapter Outline 6.1 Common and Systematic
EXPERIMENT 9: Double Replacement Reactions PURPOSE a) To identify the ions present in various aqueous solutions. b) To systematically combine solutions and identify the reactions that form precipitates
Ionic Compounds and Ionic Bonding 1 Chemical Bonds There are three basic types of bonds: Ionic The electrostatic attraction between ions Covalent The sharing of electrons between atoms Metallic Each metal
CHEM1001 2014-J-2 June 2014 22/01(a) What is the molarity of the solution formed when 0.50 g of aluminium fluoride is dissolved in 800.0 ml of water? 2 The molar mass of AlF 3 is: molar mass = (26.98 (Al)
Chemical Names & Formulas Water Ammonia Methane 1 Why Systematic Names? # atomic particles 3 (p, n, e) # elements ~120 # elements in 8 earth s crust (99%) # elements in all 25 living things # compounds
Chapter 5 - Molecules and Compounds How do we represent molecules? In pictures In formula In name Ionic compounds Molecular compounds On the course website, you will find a list of ions that I would like
1 Experiment 5: Studying Chemical Reactions When a chemical reaction occurs, substances called reactants are transformed into different substances called products that may have different appearances and
Work Session 2b: Formulas, Names, and Masses Use the information in your textbook for this work session, as well as the information on pages 4 and 5. Part 1: Binary Ionic Compounds Give the formula and
9 CHEMICAL NAMES AND FORMULAS SECTION 9.1 NAMING IONS (pages 253 258) This section explains the use of the periodic table to determine the charge of an ion. It also defines polyatomic ion and gives the
Inorganic Nomenclature Chem 1-A Lab Lecture Spring 010 Chemical Formulas: the acronyms of chemistry Composed of element symbols, subscripts and sometimes parentheses Subscripts tell the number of atoms
Created by Kristen Henry Jones 2/25/2009 AP* CHEMISTRY EQUATIONS BY TYPE Double Replacement 1. Hydrogen sulfide is bubbled through a solution of silver nitrate. 2. An excess of sodium hydroxide solution
Unit III Lecture 10 Chemistry The Molecular Nature of Matter and Change Fifth Edition Chapter 2 The Components of Matter Martin S Silberberg Copyright! The McGrawHill Companies, Inc Permission required
Objectives Name cations, anions, and ionic compounds. Write chemical formulas for ionic compounds such that an overall neutral charge is maintained. Explain how polyatomic ions and their salts are named
Chapter 8 Chemical Quantities Introductory Info The atomic masses of the elements on the periodic table are in the units. These measurements are based on the mass of the standard isotope of the element,
Single and Double Displacement Reactions Objectives To perform and observe a variety of single and double displacement reactions To record observations in detail To identify the products formed in each
CHEMICAL QUANTITIES Chapter 10 What is a mole? A unit of measurement in chemistry 1 mole of a substance = 6.02 x 10 23 (Avagadro s number) representative particles of a substance Representative particle
Elements and Compounds elements combine together to make an almost limitless number of compounds the properties of the compound are totally different from the constituent elements Tro, Chemistry: A Molecular
Chapter Exercise Key 1 Chapter Exercise Key Exercise.1 Classifying Compounds: Classify each of the following substances as either a molecular compound or an ionic compound. a. formaldehyde, CH 2 O (used
FORMULAS AND NOMENCLATURE OF IONIC AND COVALENT COMPOUNDS Adapted from McMurry/Fay, section 2.10, p. 56-63 and the 1411 Lab Manual, p. 27-31. TYPES OF COMPOUNDS Ionic compounds are compounds composed of
In the previous section, you learned how and why atoms form chemical bonds with one another. You also know that atoms combine in certain ratios with other atoms. These ratios determine the chemical formula
"Seeing" Ionic Formulas in Chemistry Dr. Malonne Davies 1, Dr. Arthur Landis 2, and Linda Landis 3 1 Department of Physical Sciences 2 Department of Physical Sciences 3 Science and Mathematics Education
The Chemistry Name Game Contributed by Butane: The Carroll College Chemistry Club Main Science Idea for Kids As students play the Chemistry Name Game, they will learn why compounds form as they do. They
CHAPTER Language of Chemistry 6 Sect 6.1 Classificat of Compounds 2. Compound Classificat Compound Classificat (a) NH 3 binary molecular (b) Fe 2 O 3 binary ic (c) BaSO 4 ternary ic (d) H 2 SO 4 (aq) ternary
Naming Compounds Handout Key p. 2 Name each of the following monatomic cations: Li + = lithium ion Ag + = silver ion Cd +2 = cadmium ion Cu +2 = copper (II) ion Al +3 = aluminum ion Mg +2 = magnesium ion
21 st Century Chemistry Multiple Choice Question in Topic 3 Metals Unit 11 1. Consider the equation: 2Ca(s) + O 2 (g) 2CaO(s) Which of the following statements are correct? (1) Calcium and oxygen are reactants.
SCH 4C1 Unit 2 Problem Set Questions taken from Frank Mustoe et all, "Chemistry 11", McGraw-Hill Ryerson, 2001 1. A small pin contains 0.0178 mol of iron. How many atoms of iron are in the pin? 2. A sample