1 / 23

Pure Substances Elements

Pure Substances Elements. Composed of one type of atom Metal, Non-metal or Metalloid Simplest type of matter that retains characteristic properties. Pure Substances Compounds. A combination of two or more elements Properties differ from those of component elements

tyler
Download Presentation

Pure Substances Elements

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. Pure SubstancesElements Composed of one type of atom Metal, Non-metal or Metalloid Simplest type of matter that retains characteristic properties

  2. Pure SubstancesCompounds • A combination of two or more elements • Properties differ from those of component elements • Separated by chemical means

  3. MixturesHeterogeneous • Has one or more visible boundaries between the components • Composition is not uniform • Some boundaries may only be seen with a microscope ie. Blood, milk

  4. MixturesHomogeneous • No visible boundaries because components are mixed as individual atoms, ions or molecules • Composition is uniform on a molecular level • Alloy- steel, a combination of Fe and C We cannot tell visually whether an object is a substance or a homogenous mixture

  5. Filtration • Separates the components of a mixture on the basis of differences in particle size • Usually a heterogeneous mixture • Most common method is to use filter paper to separate the components. The remaining liquid, after separation, is the filtrate.

  6. Separation techniques • Filtration – uses a porous barrier to separate a (non-dissolved) solid from a liquid

  7. Distillation • Separates components through differences in volatility : used for homogenous mixtures • The mixture is heated until the substance with the lowest boiling point becomes a vapor that can be condensed into a liquid and collected. • If the distillation is controlled ,you can separate substances that differ by a few degrees.

  8. Distillation – separates liquids based on their boiling points

  9. Crystallization • Based on differences in solubility • When a solution contains as much dissolved substance as it can possibly hold it has reached the saturation point • The addition of even a tiny amount causes the dissolved substance to come out and collect as crystals

  10. Crystallization – a method of extracting dissolved solids from a solution

  11. Chromatography • Based on differences in solubility • The separation is based on the various components of the mixture spreading out through the paper at different rates • Components with the strongest attraction for the paper move slower (highly soluble)

  12. Chromatography – separates components of a mixture based on how well it can travel across a particular material

  13. Law of Conservation of Mass Mass is neither created nor destroyed it just changes form: Mass reactants = Mass products

  14. A 10.0 g sample of magnesium reacts with oxygen to form 16.6 g of magnesium oxide. How many grams of oxygen reacted? • What part of the reaction is a compound?

  15. From a laboratory process designed to separate water into hydrogen and oxygen gas, a student collected 10.0 g of hydrogen and 79.4 g of oxygen. How much water was originally involved in the process?

  16. If 19.9 grams of copper are burned in air to produce 25.0 grams of copper oxide, what is the mass of oxygen from the air that is needed?

  17. Law of Conservation of Mass • Applies to compounds: The mass of a compound is equal to the sum of the masses of the elements that make up that compound

  18. 20.00 g of sugar Carbon 8.44 Hydrogen 1.30 Oxygen 10.26 sum 20.00 Find Percent by Mass of each element:

  19. Law of Definite Proportions percent by mass = mass of element x 100 mass of compound

  20. Carbon: 8.44 g C X 100 = 42.20 % 20.00 g sucrose Hydrogen: 1.30 g H X 100 = 6.50 % 20.00 g sucrose Oxygen:

  21. Analyze 500.0 g sucrose • Carbon 211.0 • Hydrogen 32.5 • Oxygen 256.5 sum 500.0

  22. A 78.0 g sample of an unknown compound contains 12.4 g of hydrogen. What is the percent by mass of hydrogen in the compound?

  23. What is the percentage by mass of I if 4.07 g reacts with 10.63 g of II to form a compound of mass 14.70 g? Also, calculate the percentage by mass of II in the compound.

More Related