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Chapter 7: Chemical Formulas and Chemical Compounds
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7.1 Chemical Names & Formulas There are MILLIONS of natural & synthetic compounds out there. Some have common names (limestone- calcium carbonate, baking soda- sodium bicarbonate), while others do not. How are we supposed to keep all these chemical straight? How are we supposed to know anything about the compound from the name? How are Scientists from around the world supposed to know what we are talking about?
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Significance of a Chemical Formula In chemical formulas, subscripts are used to give information about the compound. – Molecular compounds: subscripts indicate how many atoms of a given element. Ex. C 132983 H 211861 N 36149 O 40883 S 693 (Titin aka connectin) – Ionic compounds: the simplest ratio of cations and anions. Ex. Al 2 (SO 4 ) 3
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Monatomic Ions Ions made from a single atom. Ex. Na + or O 2- Groups 1a, 2a, 3a-7a. Groups 1a-4a form cations, 5a-7a form anions. Naming Monatomic Ions: – Cations always say their names. – Anions change their endings to IDE.
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Examples: – Na+ = sodium ion – Zn 2+ = zinc ion – Examples: O 2- = oxide Cl - = chloride
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Binary Ionic Compounds When ions bond they form ionic compounds, of which the simplest type is the binary ionic compound. – This occurs when only 2 elements, or 2 monatomic ions, bond. These compounds are named by First, writing the name of the cation followed by the name of the anion. For example… the ionic compound made of Na and F is called sodium fluoride… the ionic compound made of Zn and O is called zinc oxide. Unfortunately, these names do not tell you the ratios of the element being bonded. Is it 1:1 or 1:3? When combining cations & anions into an ionic compound, you always put the cation name first & then the anion name (the molecules are also written in that order as well.) Example: Na + + Cl - --> NaCl sodium + chloride --> sodium chloride
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Practice Compound Name CaI 2 Mg 3 P 2 LiBr Al 2 O 3
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The Stock System Nomenclature Some elements can form more than one type of cation. – Elements from the d-block can do this. The exceptions are Ag (1+), Zn(2+) and Cd (2+). The charge is indicated by the Roman numeral in parentheses followed by the word "ion“. Examples: – Fe 2+ = iron (II) ion – Fe 3+ = iron (III) ion
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Compounds Containing Polyatomic Ions Some polyatomic are oxyanions: made with O. – The most common type have an –ate ending. – The ion with 1 less O have an –ite ending. – An ion with 1 less O than –ite adds the prefix hypo- to the name. – An ion with 1 more O than –ate adds the prefix per- – Use your polyatomic ion chart
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Naming Binary Molecular Compounds The less electronegative atom is written first (the atom which is the furthest to the left and to the bottom of the periodic table) The more electronegative second atom has an "-ide" ending. A prefix indicates the number of each atom present in the compound.
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Prefixes Number of Atoms Prefix 1 Mono (don’t use on 1 st element) 2Di 3Tri 4Tetra 5Penta 6Hexa 7Hepta 8Octa 9Nona 10Deca
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Acids & Salts Binary acids- acids that are made of only two elements (no Oxygen) – A. Prefix is always hydro – B. Name the second element with the suffix- ic EX: HClHydrochloric acid
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Ternary acids- those acids that are made of more than two elements; usually contain a polyatomic ion. Oxyacids- contain hydrogen, oxygen and a 3 rd element, usually a non-metal. – A. For the acid containing the most common polyatomic ion of its group simply use the first part of the polyatomic name and follow with the suffix ic. – B. For the acid containing the polyatomic with one less oxygen than the ic, use the suffix ous. – C. For the acid containing the polyatomic with two less oxygens than the ic, use the prefix hypo and the suffix ous. – D. For an acid containing the polyatomic with one more oxygen than the ic, use the prefix per and the suffix ic.
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Examples of acid nomenclature Rule 1: all acids must contain hydrogen Rule 2: acids with `ic' suffix represent natural `ate' polyatomic ions HBrO 3 bromic acid Rule 3: when all oxygen atoms are removed, add `hydro' prefix to name HBr hydrobromic acid HCl hydrochloric acid Rule 4: when an extra oxygen is added, add a `per' prefix to name HBrO 4 perbromic acid HClO 4 perchloric acid
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5: when 1 oxygen is taken away (from `ate' ion number), change the `ic' suffix to `ous' HBrO 2 bromous acid HClO 2 chlorous acid Rule 6: when 2 oxygens are taken away (from `ate' ion number), change the `ic' suffix to `ous' and add a `hypo' prefix HBrO hypobromous acid HClO hypochlorous acid
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Salts- an ionic compound composed of a cation and the anion from an acid. Some salts contain anions in which one or more hydrogen atoms from the acid remain. These are named by adding hydrogen or the prefix bi- to the anion name. – HCO 3 - or – Hydrogen carbonate or bicarbonate
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7.2 Oxidation Numbers Oxidation Number- Indicates the general charge distribution of electrons in covalent bonds and polyatomic ions. Oxidation state-elements can have various oxidation states depending on the number of electrons that they have compared to the number of protons that are in the nucleus
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Rules for determining Ox. State 1. Free elements are assigned an oxidation state of 0. – e.g. Al, Na, Fe, H 2, O 2, N 2, Cl 2 etc have zero oxidation states. 2. The oxidation state for any simple one-atom ion is equal to its charge. – e.g. the oxidation state of Na + is +1, Be 2+, +2, and of F -, -1. 3. The alkali metals (Li, Na, K, Rb, Cs & Fr) in compounds are always assigned an oxidation state of +1. – e.g. in LiOH (Li, +1), in Na 2 SO 4 (Na, +1). 4. Fluorine in compounds is always assigned an oxidation state of -1. – e.g. in HF 2 -, BF 2 -.
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5. The alkaline earth metals (Be, Mg, Ca, Sr, Ba, and Ra) and also Zn and Cd in compounds are always assigned an oxidation state of +2. Similarly, Al & Ga are always +3. – e.g. in CaSO 4 (Ca, +2), AlCl 3 (Al, +3). 6. Hydrogen in compounds is assigned an oxidation state of +1. Exception - Hydrides, e.g. LiH (H=-1). e.g. in H 2 SO 4 (H, +1). 7. Oxygen in compounds is assigned an oxidation state of -2. Exception - Peroxide, – e.g. H 2 O 2 (O = -1). e.g. in H 3 PO 4 (O, -2).
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8. The sum of the oxidation states of all the atoms in a species must be equal to net charge on the species. – e.g. Net Charge of HClO 4 = 0, – i.e. [+1(H)+7(Cl)-2*4(O)] = 0 – Net Charge of CrO 4 2- =-2, – To solve Cr's oxidation state: x - 4*2(O) = -2, x = +6, so the oxidation state of Cr is +6.
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Oxidation Practice Au 3+ N 2 FeCl 3 N 2 O 3 NO 3 - PCl 5 OF 2 H 2 SO 4 HNO 2 MnO 4 - Au is +3 N is 0 Cl is -1 Fe is +3 O is -2 N is +3 O is -2N is +5 Cl is -1P is +5 F is -1O is +2 O is -2 H is +1S is +6 O is -2 H is +1 N is +3 O is -2Mn is +7
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Using Ox #s for Formulas & Names ( for molecules/ covalent bonds) Prefix System PCl 3 Phosphorus trichloride PCl 5 Phosphorus pentachloride N 2 O dinitrogen monoxide NO nitrogen monoxide Stock System Phosphorus (III) chloride Phosphorus (V) chloride Nitrogen (I) oxide Nitrogen (II) oxide
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7.3 Using Chemical Formulas A review from chapter 3 Formula Mass- the sum of the atomic masses of all the atoms in the compound. Ex. CO 2 C- 12.011 O- 15.999 X 2 44.009 amu – formula mass YOU TRY! 1.H 2 O 2.CH 4 Cl 2
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“The mass in grams of 1 mole of a substance..” (M). Molar Mass of… – Ca 40 amu –> 40 g/mol – C 12 amu -> 12 g/mol – H 2 O 2 34 amu -> 34 g/mol – NaCl – CaCl 2 – C 6 H 12 O 6
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Gram-mole-Particle Conversions We can use the molar mass to convert moles to grams or grams to moles. Likewise, moles can be converted to # of particles via Avogadro’s #.
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Percentage Composition If you spend 6 hours at school every day for 20 days in a month. WHAT IS THE PERCENT OF TIME you spent at school for the whole month if there are 31 days in the month? – % = (actual time ÷ total time) x 100 ((6x20) / (24 x 31)) X 100 = 16.13%
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What does this have to do with formulas? Similarly, we can calculate the percents of element mass in a compound. “The mass of each element in a compound compared to the entire mass of the compound and multiplied by 100 percent is called the percentage composition of the compound.” %= (mass of element ÷ total mass of the compound) x 100
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For example… – If you have one mole of water (total mass of 18g), what is the percent composition of H and O? Hint: You need to know the formula to solve for % comp if you don’t know the mass for each element. H 2 O- which means you have 2 mol H and 1 mol O. Well… what is the mass of 2 mole H and what is the mass of 1 mol of O? H – 2g & O- 16g Now… calculate… (mass of element ÷ total mass) x 100 – % H = (2/18)x 100 = 11% – % O = (16/18) x 100 = 89%
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7.4 Determining Chemical Formula Empirical Formula: “A formula that gives the simplest whole- number ratio of the elements is called an empirical formula.” We can use the ratio of the masses to determine this formula.
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Working the H 2 O example backwards If we know we have 11% H and 89% O what would the EMPIRICAL formula for water be? 1. Pretend you have a 100g sample. CHANGE % to g. 2. Convert grams to moles 3. Divide each mole by the SMALLEST mole #. 4. Round to a whole number and write the empirical formula – H: 11% of 100g = 11g – O: 89% of 100g = 89g – H: 11 g /1 g = 11 mol H – O: 89 g /16 g = 5.56 mol O – H: 11/5.56 = 1.977 – O: 5.56/5.56 = 1 – H: 2 – O: 1 –H2O–H2O
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Molecular Formula “The formula that gives the actual number of atoms of each element in a molecular compound is called the molecular formula.” This will be the EMPIRICAL formula multiplied by a whole number. – Ex. OH x 2 = H 2 O 2 … hydrogen peroxide.
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How do you know what whole # to multiply the empirical formula by? The question will always tell you the MOLAR MASS of the molecular compound. Take the molar mass of the compound and divide by the empirical formula mass. Molar mass ÷ Empirical formula mass The answer is the WHOLE # used to multiply the empirical formula by = molecular formula.
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150 g/mol EX: Ribose has a molar mass of 150 g/mol and a chemical composition of 40.0% C, 6.67% H and 53.3% O. What is the molecular formula? – First, find the empirical formula. 1. Pretend you have a 100g sample. 2. Convert grams to moles 3. Divide each mole by the SMALLEST mole #. 4. Round to a whole number and write the empirical formula – Now calculate the formula mass of the empirical formula. – Divide molar mass by formula mass. Molar mass ÷ Empirical formula mass – Multiply the whole # by the empirical formula = molecular formula.
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