Chem 20A Homework 1 KEY 2 Absolute alcohol Substance compound Milk mixture homogeneous Copper wire substance element Rust mixture homogeneous or substance compound Barium bromide substance compound Concrete mixture heterogeneous Baking soda substance compound Baking powder mixture homogeneous 4 In the 1800 s scientific technology was not as advanced as it was today making it difficult for scientists to completely characterize isolated materials Undeveloped methods of physical separation and characterization resulted in false reports In more recent times the inability to confirm new elements lies in the nature of how new elements are created Since all naturally occurring elements have been discovered and isolated new elements must be formed by colliding known elements making them very unstable and short lived Due to these issues they are difficult to characterize and confirm as elements 6 25 0 g Hf 0 125 g Hf 0 125 g Hf 31 5 g Te x x 0 1575 g Te 31 5 g Te x g Te 25 0 g Hf 8 a Using the Law of constant composition assume you have 100 g of each binary compound to solve for the amount of fluorine when amount of iodine equals 1 g Compound 1 1 000 g I 13 021 g F 0 1497 g F 86 979 g I Compound 2 1 000 g I 30 993 g F 0 4491 g F 69 007 g I Compound 3 1 000 g I 42 809 g F 0 7485 g F 57 191 g I Compound 4 1 000 g I 51 171 g F 1 0479 g F 48 829 g I b Find small whole number ratios by dividing each value by the smallest value Compound 1 0 1497 1 0 1497 Compound 2 0 4491 3 0 1497 Compound 3 0 7485 5 0 1497 Compound 4 1 0479 7 0 1497 Yes these compounds satisfy the Law of Multiple proportions 12 N2H4 liq N2 g 2 H2 g From the equation we know for every 1 volume of N2 there are 2 volumes of H2 produced Therefore 1 vol N2 13 7 mL N2 x 2 13 7 mL 27 4 mL H2 2 vol H2 x mL H2 14 2 CH3OH g 3 O2 g 4 H2O g 2 CO2 g Since all components are in the gas state the equation can be re written as follows 2 volumes CH3OH 3 volumes O2 4 volumes H2O 2 volumes CO2 4 volumes H2O x L H2O x 4 0 L H2O 2 volumes CH3OH 2 L CH3OH 2 vol CO2 x L CO2 x 2 0 L CO2 2 volumes CH3OH 2 L CH3OH 15 27 97693 x 92 21 100 28 97649 x 4 70 100 29 97376 x 3 09 100 28 08561 28 086 22 where A mass number and Z number of protons In a neutral compound protons electrons and A protons neutrons Therefore protons 109 electrons 109 neutrons 266 109 157 24 a In order to determine the charge of an electron based on the experimental data given in this problem we need to find the smallest value that the charge increases by aka the lowest common denominator which we will assume is the charge of the electron since this is the only data we are given Following the hint given in the question we first need to find the difference in the entries so we can see by how much the charge increases each time Looking at these values we can see they are either 1 6 x 10 19 or 3 2 x 10 19 2 1 6 x 10 19 From these values we are able to determine the lowest common denominator as approximately 1 6 x 10 19 meaning this the experimentally determined approximate charge of the electron which can be used to find the number of electrons in each oil drop Charge of Oil Drop Difference from previous Number of electrons in entry each oil drop 19 6 563 x 10 4 19 19 8 204 x 10 1 641 x 10 5 11 50 x 10 19 3 296 x 10 19 7 19 19 13 13 x 10 1 63 x 10 8 16 48 x 10 19 3 35 x 10 19 10 18 08 x 10 19 1 6 x 10 19 11 19 19 19 71 x 10 1 63 x 10 12 22 89 x 10 19 3 18 x 10 19 14 19 19 26 18 x 10 3 29 x 10 16 b Charge of the electron 1 6 x 10 19 C c Yes it is possible that the charge of the electron is smaller than the value determined above Since we were using the experimental data and were given no additional information we had to assume the smallest increase in total charge came from increasing by 1 electron but it is possible that this increase was from 2 3 4 etc electrons In order to prove this is the actual charge more data should be collected 27 Dalton s Original Postulates 1 Matter consists of indivisible atoms Correction Atoms are not indivisible Such as atoms can lose or gain electrons to give ions with unique properties Or radioactive elements that spontaneously emit or absorb subatomic particles 2 All atoms of a given chemical element are identical in mass and all other properties Correction Isotopes of elements exist They have virtually all the same properties but isotopic effects can be observed They also differ in weight due to the difference in neutrons 3 Different chemical elements have different kinds of atoms and such atoms have different masses No change 4 Atoms are indestructible and retain their identity in chemical reactions Correction Atoms are not indestructible they can be split apart or fused to give new kinds of atoms 5 Formation of a compound from its elements occurs through combining atoms of unlike elements in small whole number ratios Correction Certain solid compounds vary in a range as seen in non stoichiometric compounds The law of definite proportions is always true for gaseous and liquid compounds but not always for solids