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Empirical and Molecular Formulas How to find out what an unknown compound is

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A chemist obtains a new product What is the formula for the compound? First step – determine constituent elements and their amounts This info can be used to determine chemical formula

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Formula of a compound represents relative numbers of atoms present E.g. “CO 2 ” tells us that in a molecule of this compound there is 1 carbon atom to every 2 oxygen atoms To determine the formula of a substance we need to count the atoms – we can do this by weighing

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An example You have a compound that you know contains only Carbon, Hydrogen and Oxygen. You have a 0.2015 g sample Analysis shows you have 0.0806 g C, 0.01353 g H, and 0.1074 g O We can convert these masses to moles, and then moles to atoms using dimensional analysis…..

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Determining # of moles of elements in unknown compound

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Determining number of atoms of unknown compound

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To summarize up to this point We have.00671 moles of Carbon We have.01342 moles of Hydrogen We have.006713 moles of Oxygen Amount of Carbon = Amount of Oxygen .01342/.006713 = 2, so we have twice as much Hydrogen as we have of Oxygen and as we have of Carbon We have a ratio of 1:2:1

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We know the ratio of elements 1 Carbon : 2 Hydrogens : 1 Oxygen We can write this as CH 2 O Is this the molecular formula? Maybe…but the molecule might also have 2 Carbons, 4 Hydrogens and 2 Oxygens, or it might have 16 Carbons, 32 Hydrogens and 16 Oxygens We have found the EMPIRICAL FORMULA – a formula that represents the ratio of elements in a compound. This is also called “simplest formula” since it is smallest whole-number ratio of elements in the compound

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Empirical Formula Vs Molecular Formula Empirical formula gives relative numbers of atoms e.g. CH 2 O Molecular formula gives the actual numbers of atoms e.g. C 6 H 12 O 6 C 6 H 12 O 6 = (CH 2 O) 6

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Find the Empirical Formula Empirical Formula of Benzene = CH Empirical Formula of Dioxin = C 6 H 2 Cl 2 O

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How to calculate empirical formula We have 0.2636 grams of nickel. We heat it in the presence of oxygen to produce 0.3354 grams of a nickel oxide. What is the formula of the compound we made? First - what is the mass of oxygen that reacted with the copper? Mass of copper oxide – Mass of copper = mass of oxygen 0.3354 g copper oxide – 0.2636 g copper = 0.0718 g oxygen Next – Find number atoms involved…

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Find number of atoms in the compound Mole quantities represent number of atoms We have same number of moles of oxygen as of nickel Empirical formula will be NiO

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Another example You have a metal oxide made by reacting 4.151g Al with 3.692 g O. What is the empirical formula? Atomic mass Al = 26.98 g/mol Atomic mass O = 16.00 g/mol Need to know relative numbers of atoms, so need to convert grams to moles and then find the whole number ratio of atoms

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Find whole number ratios Do this by dividing both numbers by the smallest of the two. This converts the smallest number to 1 This is not a whole number ratio – to get to a whole number, all we have to do is multiply by 2

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Finally we have found our empirical formula Al 2 O 3

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Summary

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If you have percent composition your can also find the empirical formula You have a compound that is 27% Carbon and 73% Oxygen by mass. What is the empirical formula? Assume you have 100 grams. That means you would have 27 g C and 73 g O Convert grams to moles: 27 g C (1 mol/12 g C) = 2.25 mol C 73 g O (1 mol/16 g O) = 4.6 mol O Divide by smallest # of moles: 4.6 /2.25 = 2.04 (O) 2.25/2.25 = 1 (C) Empirical formula = CO 2

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A poem to help Percent to mass Mass to mole Divide by small Multiply ‘til whole

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Find empirical formula from percent composition Nylon-6 is a compound that is 63.68 C, 12.38% N, 9.80% H and 14.4% O. Find the empirical formula C 6 H 11 NO

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One more step to find molecular formula To find the molecular formula of your unknown compound, you need another piece of information – the molar mass of the compound, in addition to the percent composition

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You have the empirical formula – what is the molecular formula? Empirical formula = P 2 O 5 Molar mass of compound (determined experimentally) is 238.88 g/mol P = 30.97 g/mol ; O = 16.00 g/mol Find empirical formula mass - the molar mass that the compound would have based on the empirical formula: 2(30.97g/mol) + 5(16.00 g/mol) = 141.94 g/mol

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Finding molecular formula The molar mass of the compound will be some multiple of the empirical mass – find the multiple by dividing molar mass by empirical mass: (238.88 g/mol)/(141.94 g/mol) = 1.7 1.7 ≈ 2 Molar mass of unknown is about twice empirical mass. We will multiply empirical formula subscripts by 2 Molecular formula = P 4 O 10

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One last problem

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Homework Read Ch 6, section 3 (pp. 196 – 208) Do empirical formula homework sheets Do molecular formula homework sheets Monday – we will go over homework and talk about hydrates Test Review Test on Wednesday!!!! We have 12 class meetings to go until semester final

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