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Empirical Formulas. CH 3. Empirical Formula. Empirical Formula- Smallest whole number ratio of atoms in a compound. C 2 H 6. reduce subscripts. May not be the actual molecular formula. To Find Empirical Formula. 1. Find mass (or %) of each element . 2. Find moles of each element.
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CH3 Empirical Formula EmpiricalFormula-Smallest whole number ratio of atoms in a compound C2H6 reduce subscripts May not be the actual molecular formula.
To Find Empirical Formula 1. Find mass (or %) of each element. 2. Find moles of each element. 3. Divide moles by the smallest # to find subscripts. 4. When necessary, multiply subscripts by 2, 3, or 4 to get whole #’s. Memory Tool Next
Use: % to mass Mass to mole Divide by small Multiply ‘til whole
Empirical Formula-Together • Qualitative analysis shows that a compound contains 32.38% Na, 22.65%S, and 44.99% O. Find the empirical formula of this compound. Answer next slide
Follow the steps! 32.38 g Na x 1 mol Na = 1.408 mol Na = 1.993= 2 22.99 g Na 0.7063 22.65 g S x 1 mol S = 0.7063 mol S = 1 32.07 g S 0.7063 44.99 g O x 1 mol O = 2.812 mol O = 3.981 = 4 16.00 g O 0.7063 **empirical formula for this compound is: Na2SO4
Empirical Formula • Find the empirical formula for a sample of 25.9% N and 74.1% O. Try on your own. Answer next slide.
Empirical Formula • Find the empirical formula for a sample of 25.9% N and 74.1% O. 25.9 g N x 1 mol N = 1.85 mol N = 1 N 14.01 g N 1.85 mol 74.1 g O x 1 mole O = 4.63 mol O = 2.5 mol O 16.00 g O 1.85 mol So the Empirical Formula is???
Empirical Formula N1O2.5 Need to make the subscripts whole numbers multiply by 2 Empirical formula for this compound is N2O5
Numbers to watch out for after dividing by the smallest number of moles (don’t need to copy, just be aware of) Number what to multiply by • 0.25 4 • 0.33 3 • 0.5 2 • 0.66 3 • 0.75 4
C2H6 Molecular Formula “True Formula”, the actualnumber of atoms in a compound CH3 empirical formula ? molecular formula
Calculating Molecular Formula 1. Find the empirical formula. 2. Find the empirical formula mass. 3. Divide the molecular mass by the empiricalmass. 4. Multiply each subscript by the answer from step 3.
Molecular Formula 1) Molecular Form Mass= X Empirical Form Mass 2) Then multiply X by subscripts in empirical formula
Calculating Molecular Formula • The empirical formula for ethylene is CH2. Find the molecular formula if the molecular mass is 28.1 g/mol? Empirical formula mass = 14.03 g/mol 28.1 g/mol 14.03 g/mol = 2.00 = (CH2)2 C2H4
Calculating Molecular Formula • If 4.04 g of N combine with 11.46 g O to produce a compound with a formula mass of 108.0 amu, what is the molecular formula of this compound? • !st you must find the Empirical Formula
4.04 g N x 1mol N = .288 mol N Divide by small .288/.288 N=1 11.46 g O x 1 mol O = .7163 mol O Divide by small .7163/.288 O= 2.49 Multiply ‘til whole: N x 2 O x 2
Calculating Molecular Formula Empirical Formula (N1O2.5)2 = N2O5
Calculating Molecular Formula The Molar mass of a compound is 92 g/mol. Analysis of a sample of the compund indicates that is contains 0.606 g N and 1.390 g O. Find its molecular formula.
.606 g N x 1 mol N = .0436 mol N = 1 14.01 g N .0436 1.390 g O x 1 mol O = .08688 mol O = 1.993 16.00 O .0436 You can round up the 1.993 Go the next slide
Calculating Molecular Formula Empirical form = NO2 Molecular mass = 92 = 2 Empirical mass 46 (NO2)2 = N2O4
C2H6 Comparison of the Two empirical formula CH3 molecular formula