1 / 20

Molar Mass & Percent Composition

Molar Mass & Percent Composition. Moles. How many moles are present? CH 4 C = 1 mol , H = 4 mol. Molar Mass. Molar mass of any substance is the mass (in grams) of 1 mol of the substance. The molar mass is obtained by summing the masses of the component atoms. CH 4

keithrkelly
Download Presentation

Molar Mass & Percent Composition

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Molar Mass &Percent Composition

  2. Moles • How many moles are present? CH4 C = 1 mol, H = 4 mol.

  3. Molar Mass Molar mass of any substance is the mass (in grams) of 1 mol of the substance. The molar mass is obtained by summing the masses of the component atoms. CH4 Mass of 1 mol of C = 1 x 12.0 g = 12.0 g Mass of 4 mol of H = 4 x 1.0 g = 4.0 g Mass of 1 mol of CH4 = 16.0 g

  4. Molar Mass H2O Mass of 2 mol of H = 2 x 1.0 g = 2.0 g Mass of 1 mol of O = 1 x 16.0 g = 16.0 g Mass of 1 mol of H2O = 18.0 g

  5. Percent Composition The % mass that one particular element contributes to an entire compound. Mass percent = mass element present x 100 mass compound • to calculate… ~ find total mass of one molecule ~÷ massofelement (part)bytotal mass (whole) ~ x by 100

  6. Example • What is the Mass Percent of Na2S? • Step 1 find Molar Mass. • Na = 2 x 23.0 = 46.0 g • S = 1 x 32.0 = 32.0 g • Molar Mass = 78.0g • Step 2 divide individual elements mass by the molar mass of entire compound. Sodium = (46.0 g / 78.0 g) x 100 = 59.0% Sulfur = (32.0g / 78.0g) x 100 = 41.0%

  7. Practice #1 • Calculate the mass percent of C10H14O. • Carbon = 10 x 12.0g = 120.0g • Hydrogen = 14 x 1.0g = 14.0g • Oxygen = 1 x 16.0g = 16.0g • Molar mass = 150.0g

  8. Practice #1 Mass Carbon = 120.0g /150.0g x 100 = 80.0% Mass Hydrogen = 14.0g/150.0g x 100 = 9.3% Mass Oxygen = 16.0g / 150.0 g x 100 = 10.7%

  9. Practice #2 • Determine mass percent of sulfuric acid. • H4(SO4)2 • Hydrogen = 4 x 1.0g = 4.0g • Sulfur = 2 x 32.1g = 64.2g • Oxygen = 8 x 16.0g = 128.0g • Molar Mass = 196.2g

  10. Practice #2 Mass Hydrogen = 4.0g / 196.2g x 100 = 2.04% Mass Sulfur = 64.2g / 196.2g x 100 = 32.7% Mass Oxygen = 128.0g / 196.2g x 100 = 65.2%

  11. Practice #3 • Determine mass percent of sulfuric acid. • Ca(NO3)2 • Calcium = 1 x 40.1g = 40.1g • Nitrogen = 2 x 14.0g = 28.0g • Oxygen = 6 x 16.0g = 96.0g • Molar Mass = 164.1g

  12. Practice #3 Mass Calcium = 40.1g / 164.1g x 100 = 24.4% Mass Nitrogen = 28.0g / 164.1g x 100 = 17.1% Mass Oxygen = 96.0g / 164.1g x 100 = 58.5%

  13. Empirical formula---lowest ratio of elements in compound Molecular formula ---actual ratio of elements in compounds Subscripts are mole ratios EMPIRICAL AND MOLECULAR FORMULAS

  14. Compounds may have the same empirical formula but different molecular formulas. For example CH is the empirical formula for C2H2 and C6H6 If the subscripts can be reduced, then the formula is a molecular formula.

  15. Empirical Formulas • The lowest whole # ratio of a compound. (reduce the subscripts down if you can) • Ex. the empirical formula for glucose (C6H12O6) would be… CH2O • to calculate from % composition… 1. ÷ given % by atomic mass (to find moles) 2.÷ each mole in step 1 by lowest mole value (to find ratio) 3. values from step 2 = subscripts ~subscripts MUST be a whole # (multiply if they are not!)

  16. Empirical Formulas • practice… 1.What is the empirical formula of a compound that contains 36.5% sodium, 25.4% sulfur and 38.1% oxygen? Na=36.5% =1.59 moles =2 23g 0.794 Na SO 2 3 S=25.4% =0.794 moles *** =1 subscripts 32g 0.794 O=38.1% =2.38 moles =3 16g 0.794

  17. Empirical Formulas • more practice… 2.What is the empirical formula of a compound that contains 72.4% iron (Fe) and 27.6% oxygen? Fe=72.4% =1.293 moles *** =1 x 3 = 3 56 1.293 O=27.6% =1.725 moles x 3 = 4 =1.33 16 1.293 MUST be a whole #!!! Fe O 4 3

  18. You must be given the actual molecular molar mass of the compound in the problem. Divide the actual molecular mass by the empirical formula mass. Then multiply all the subscripts by your answer. For example the actual molar mass of the previous compound is 88.0g. The empirical formula mass is 44.0g FINDING MOLECULAR FORMULAFROM EMPIRICAL FORMULA

  19. 88.0g/44.0g = 2 C2H4O X 2 = C4H8O2

  20. H2SO4 C6H12O6 CH2O NaCl C6H12 P4O10 empirical molecular empirical empirical molecular molecular LEARNING CHECK!!!Which of the following are empirical formulas and which are molecular?

More Related