1 / 11

Chapter 3

Chapter 3 . Mass Relationships in Chemistry; Stoichiometry. Mass. Atomic masses are relative, representing the mass of an atom of one element compared to the mass of another. Carbon-12 is the standard and its mass is assigned a value of 12 amu. Atomic Mass and Isotopic Abundance.

steve
Download Presentation

Chapter 3

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. Chapter 3 Mass Relationships in Chemistry; Stoichiometry

  2. Mass Atomic masses are relative, representing the mass of an atom of one element compared to the mass of another. Carbon-12 is the standard and its mass is assigned a value of 12 amu.

  3. Atomic Mass and Isotopic Abundance Silicon consists of three stable isotopes whose atomic masses and abundances are as follows: Si-28 27.98 amu 92.2% Si-29 28.98 amu 4.7% Si-30 29.97 amu 3.1% Answer: 28.1 amu

  4. % Composition Percent composition = total mass of element/total mass of compound X 100

  5. Empirical Formula • Determine the mass of each element. • Determine the number of moles of each element by dividing by the atomic mass of that element • Find smallest number of moles. • Divide the number of moles of each element by the smallest number of moles • Number of moles are subscripts.

  6. Examples A sample of compound is made up of 78.20 g of potassium and 32.06 g sulfur. What is the empirical formula? 78.20 g K x 1 mol K/39.10 g K = 2.000 mol K 32.06 g S x 1 mol S/32.06 g S = 1.000 mol S Divide each by 1.000 K2 S1 Formula: K2S

  7. A compound has the following percent composition. 31.9 % K 29.0 % Cl 39.1% O To find the mass of each element, assume 100.0 g of compound. 31.9 g K 29.0 gCl 39.1 g O 31.9 g K x 1 mol K/39.10 g K = 0.816 mol K 29.0 gClx 1 mol Cl/35.35 gCl = 0.818 mol Cl 39.1 g O x 1 mol O/16.00 g O = 2.44 mol O Ratios: K 0.816/0.816 = 1 Cl 0.818/0.816 = 1 O 2.44/0.816 = 2.99 Simplest formula: KClO3

  8. Rounding to determine integer ratios • To get an integer, it is ok to round down by at most 0.1. • 2.09 can be rounded down to 2 • To get an integer, it is ok to round up by at most 0.1. • 2.93 can be rounded up to 3 • If that is not possible, multiply each ratio by small whole numbers until multiplication by one small number yields ratios that are all integers. • 1.50 x 2 = 3 1.33 x 3 = 4

  9. Molecular Formulas Multiple: molar mass/simplest formula mass

More Related