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pashok25 [27]
3 years ago
15

A compound is found to contain 6.360 % silicon, 36.18 % bromine, and 57.46 % iodine by mass. what is the empirical formula for t

his compound?
Chemistry
1 answer:
Karolina [17]3 years ago
7 0
<span>Percentage of the compound,
 silicon = 6.360 %
 bromine = 36.18 %
 iodine = 57.46 %
 In 100 grams of compound we will have silicon = 6.36, bromine = 36.18, iodine = 57.46
 Molar mass
 silicon = 28.09, bromine = 79.90, iodine = 126.09
Calculating the moles,
 silicon = 6.36 / 28.09 = 0.226
 bromine = 36.18 / 79.90 = 0.452
 iodine = 57.46 / 126.09 = 0.455
 Divide with the lowest
 silicon = 0.226 / 0.226 = 1
 bromine = 0.452 / 0.226 = 2
 iodine = 0.455 / 0.226 = 2
 So the empirical formula would be SiBr2I2</span><span />
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1) You need to use the atomic mass of copper.


You can find it in a periodic table. It is 63.546 amu.


2) The atomic mass is the weigthed mass of the different isotopes.


This is, the atomic mass of one element is the atomic mass of each isotope times its corresponding abundance:


=> atomic mass of the element = abundance isotope 1 * atomic mass isotope 1 + abundance isotope 2 *  atomic mass isotope 2 + ....+abundance isotope n * atomic mass isotope n.


3) The statement tells there are two isotopes so the abundance of one is x and the abundance of the other is 1 - x


=> 63.546 amu = x * 62.9296 amu + (1-x)*64.9278


=> 63.546 = 62.9296x + 64.9278 - 64.9278x


=> 64.9278x - 62.9296 = 64.9278 - 63.546


=> 1.9982x = 1.3818


=> x = 1.3818 / 1.9982 = 0.6915 = 69.15%


=> 1 - x = 1 - 0.6915 = 0.3085 = 30.85%


Answer:


Cu-63 69.15%;


Cu-65 : 30.85%
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