Download presentation

Presentation is loading. Please wait.

1
3/27/2017

2
**Simplest formula calculations**

3/27/2017 Simplest formula calculations Q- a compound is found to contain the following % by mass: 69.58% Ba, 6.090% C, 24.32% O. What is the simplest (i.e. empirical) formula? Step 1: imagine that you have 100 g of the substance. Thus, % will become mass in grams. E.g % Ba becomes g Ba. (Some questions will give grams right off, instead of %) Step 2: calculate the # of moles (mol = g ¸ g/mol) Step 3: express moles as the simplest ratio by dividing through by the lowest number. Step 4: write the simplest formula from mol ratios.

3
**Simplest formula: sample problem**

3/27/2017 Q % Ba, 6.090% C, 24.32% O. What is the empirical (a.k.a. simplest) formula? 1: g Ba, g C, g O 2: Ba: g ¸ g/mol = mol Ba C: g ¸ g/mol = mol C O: g ¸ g/mol = mol O 3: mol (reduced) mol Ba C O 1.520 / = 1 / = 1.001 1.520/ = 3.000 4: the simplest formula is BaCO3

4
**Mole ratios and simplest formula**

3/27/2017 Given the following mole ratios for the hypothetical compound AxBy, what would x and y be if the mol ratio of A and B were: A = 1 mol, B = 2.98 mol A = mol, B = 1 mol A = 2.34 mol, B = 1 mol A = 1 mol, B = 1.48 mol AB3 A4B3 A7B3 A2B3 A compound consists of 29.1 % Na, 40.5 % S, and 30.4 % O. Determine the simplest formula. A compound is composed of 7.20 g carbon, 1.20 g hydrogen, and 9.60 g oxygen. Find the empirical formula for this compound Try questions on page 189.

5
**For instructor: prepare molecular models**

Question 1 3/27/2017 1: Assume 100 g: 29.1 g Na, 40.5 g S, 30.4 g O 2: Na: g ¸ g/mol = mol Na S: g ¸ g/mol = mol S O: g ¸ g/mol = 1.90 mol O 3: mol (reduced) mol Na S O 1.266 1.263 1.90 1.266/ = 1.00 1.263/ = 1 1.90/ = 1.50 4: the simplest formula is Na2S2O3 For instructor: prepare molecular models

6
Question 2 3/27/2017 1: 7.20 g C, 1.20 g H, 9.60 g O 2: C: g ¸ g/mol = mol C H: g ¸ 1.01 g/mol = mol H O: 9.6 g ¸ g/mol = 0.60 mol O 3: mol (reduced) mol C H O 0.5995 1.188 0.60 0.5995/ = 1 1.188/ = 1.98 0.60/ = 1.0 4: the simplest formula is CH2O

7
**Question 3 1: Assume 100 g: 28.9 g K, 23.7 g S, 47.7 g O**

3/27/2017 1: Assume 100 g: 28.9 g K, 23.7 g S, 47.7 g O 2: C: 7.20 g ¸ g/mol = mol C H: 1.20 g ¸ 1.01 g/mol = mol H O: 9.6 g ¸ g/mol = 0.60 mol O 3: mol (reduced) mol C H O 0.5995 1.188 0.60 0.5995/ = 1 1.188/ = 1.98 0.60/ = 1.0 4: the simplest formula is CH2O

8
**Molecular formula calculations**

3/27/2017 There is one additional step to solving for a molecular formula. First you need the molar mass of the compound. E.g. in Q2, the molecular formula can be determined if we know that the molar mass of the compound is 150 g/mol. First, determine molar mass of the simplest formula. For CH2O it is 30 g/mol ( ). Divide the molar mass of the compound by this to get a factor: 150 g/mol ¸ 30 g/mol = 5 Multiply each subscript in the formula by this factor: C5H10O5 is the molecular formula. (models) Q- For OF, give the molecular formula if the compound is 70 g/mol O2F ¸ 35 = 2

9
**Combustion analysis gives the following: **

3/27/2017 Combustion analysis gives the following: 26.7% C, 2.2% hydrogen, 71.1% oxygen. If the molecular mass of the compound is g/mol, determine its molecular formula. What information must be known to determine a) the empirical formula of a substance? b) the molecular formula of a substance? A compound’s empirical formula is CH, and it weighs 104 g/mol. Give the molecular formula. A substance is decomposed and found to consist of 53.2% C, 11.2% H, and 35.6% O by mass. Calculate the molecular formula of the unknown if its molar mass is 90 g/mol.

10
**Question 7 1: Assume 100 g total. Thus:**

3/27/2017 1: Assume 100 g total. Thus: 26.7 g C, 2.2 g H, and 71.1 g O 2: C: g ¸ g/mol = mol C H: 2.2 g ¸ 1.01 g/mol = 2.18 mol H O: g ¸ g/mol = mol O 3: 2.223 2.18 4.444 2.223/2.18 = 1.02 2.18/ = 1 4.444/ = 2.04 4: the simplest formula is CHO2 5: factor = 90/45=2. Molecular formula: C2H2O4

11
3/27/2017 Question 8, 9 For the empirical formula we need to know the moles of each element in the compound (which can be derived from grams or %). For the molecular formula we need the above information & the molar mass of the compound Molar mass of CH = 13 g/mol Factor = 104 g/mol ¸ 13 g/mol = 8 Molecular formula is C8H8

12
**Question 10 1: Assume 100 g total. Thus:**

3/27/2017 1: Assume 100 g total. Thus: 53.2 g C, 11.2 g H, and 35.6 g O 2: C: g ¸ g/mol = mol C H: g ¸ 1.01 g/mol = mol H O: g ¸ g/mol = mol O 3: 4.430 11.09 2.225 4.43/ = 1.99 11.09/ = 4.98 2.225/2.225 = 1 4: the simplest formula is C2H5O 5: factor = 90/45=2. Molecular formula: C4H10O2

13
3/27/2017 Assignment Calculate the percentage composition of each substance: a) SiH4, b) FeSO4 Calculate the simplest formulas for the compounds whose compositions are listed: a) carbon, 15.8%; sulfur, 84.2% b) silver,70.1%; nitrogen,9.1%; oxygen,20.8% c) K, 26.6%; Cr, 35.4%, O, 38.0% The simplest formula for glucose is CH2O and its molar mass is 180 g/mol. What is its molecular formula?

14
**percentage composition molar mass(g/mol) 64.9% C, 13.5% H, 21.6% O 74 **

Determine the molecular formula for each compound below from the information listed. substance simplest formula molar mass(g/mol) a) octane C4H9 114 b) ethanol C2H6O 46 c) naphthalene C5H4 128 d) melamine CH2N2 126 The percentage composition and approximate molar masses of some compounds are listed below. Calculate the molecular formula of each percentage composition molar mass(g/mol) 64.9% C, 13.5% H, 21.6% O 74 39.9% C, 6.7% H, 53.4 % O 60 40.3% B, 52.2% N, 7.5% H 80 3/27/2017

15
**the simplest formula is CS2**

3/27/2017 1 a) Si= 87.43% (28.09/32.13 x 100), H= 12.57% b) Fe= 36.77% (55.85/ x 100), S= 21.10% (32.06/ x 100), O= 42.13% 2 a) Assume 100 g. Thus: 15.8 g C, 84.2 g S. C: 15.8 g ¸ g/mol = mol C S: 84.2 g ¸ g/mol = mol S 2.626/1.315 = 2.00 1.315/1.315 = 1 Mol reduced 2.626 1.315 Mol S C the simplest formula is CS2

16
2 b) Ag: 70.1 g ¸ g/mol = mol Ag N: 9.1 g ¸ g/mol = mol N O: 20.8 g ¸ g/mol = 1.30 mol O 3/27/2017 .6499/.6495 = 1.0 0.6499 Ag 1.30/.6495 = 2.00 .6495/.6495 = 1 Mol reduced 1.30 0.6495 Mol O N AgNO2 2 c) K: 26.6 g ¸ g/mol = mol K Cr: g ¸ g/mol = mol Cr O: 38.0 g ¸ g/mol = mol O .6803/.6803 = 1 0.6803 K 2.375/.6495 = 3.49 .6808/.6803= 1.00 Mol reduced 2.375 0.6808 Mol O Cr K2Cr2O7

17
3/27/2017 3 C6H12O6 (CH2O = 30 g/mol, 180/30 = 6) 4 a) C8H18 (C4H9 = 57 g/mol, 114/57 = 2) b) C2H6O (C2H6O = 46 g/mol, 46/46 = 1) c) C10H8 (C5H4 = 64 g/mol, 128/64 = 2) d) C3H6N6 (CH2N2 = 54 g/mol, 126/42 = 3) 5 a) C: 64.9 g ¸ g/mol = mol C H: g ¸ 1.01 g/mol = mol H O: 21.6 g ¸ g/mol = 1.35 mol O 5.404/1.35 = 4.00 5.404 C 1.35/1.35 = 1 13.37/1.35 = 9.90 Mol reduced 1.35 13.37 Mol O H C4H10O C4H10O (C4H10O = 74 g/mol, 74/74 = 1)

18
**For more lessons, visit www.chalkbored.com**

5 b) C: 39.9 g ¸ g/mol = mol C H: 6.7 g ¸ 1.01 g/mol = 6.63 mol H O: 53.4 g ¸ g/mol = mol O 3/27/2017 3.322/3.322 = 1 3.322 C 3.338/3.322 = 1.00 6.63/3.322 = 2.0 Mol reduced 3.338 6.63 Mol O H CH2O C2H4O2 (CH2O = 30 g/mol, 60/30 = 2) 5 c) 3.728/3.726 = 1.00 3.728 B 7.43/3.726 = 2.0 3.726/3.726 = 1 Mol reduced 7.43 3.726 Mol H N B3N3H6 (BNH2 = g/mol, 80/26.84= 2.98) For more lessons, visit

Similar presentations

OK

Copyright © Action Works 2008 All Rights Reserved - Photos by David D

Copyright © Action Works 2008 All Rights Reserved - Photos by David D

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on applied operational research journal Ppt on art and craft movement pottery Ppt on object oriented technologies Ppt on viruses and bacteria lesson Ppt on charge coupled device radiography Ppt on teachers day Ppt on kingdom monera pdf Ppt on transport and communication Ppt on water activity testing Ppt on non agricultural activities and pollution