Chapter 5 Chemistry Notes

# Example 90 g of water can be separated into 10 g of

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Unformatted text preview: e: 9.0 g of water can be separated into 1.0 g of hydrogen and 8.0 g of oxygen. (Note: mass conserved). Two ways to think about this: As a fraction (or %): Oxygen=8/9 (89%) Hydrogen=1/9 (11%) If we had an 18 g sample Oxygen=8/9*18g=16 g Hydrogen=1/9*18g=2 g As a proportion (O:H ratio): Since water is a pure compound (1) water is chemically the same even if the sample is very small (2) and law of constant composition should still apply to a small sample. *** ***Dalton’s theory says that this is true until we get to the atomic level. At some point we can’t get a smaller size and still be water. Macroscopic mass O 8g = =8 H 1g 1g H If a second sample had 2 g of hydrogen: O 8g Xg = = H 1g 2g X= 16 g of oxygen 8 g Oxygen Sub-microscopic masses too 1 8 If we know the atomic mass of H (1 amu) and O (16 amu), we can find out the ratio of the atoms of each type. (recall: 1 amu = 1.6x10-24g) H’s=1g/1.6x10-24g=6x1023 O’s=8g/(16*1.6x10-24g)=3x1023 **Atom ratio: 2 H per 1 O! (**This is the relative number...
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## This note was uploaded on 02/04/2014 for the course CHEM 004:007 taught by Professor Russellarsen during the Fall '13 term at University of Iowa.

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