Answer:
isotopes, there elements with a diffrent type atomic weight then the original, usally due to a higher amout of neutrons than the original. Some Isotopes are just as useable as the normal versions, but in some cases, such as Uranium, they can be even more radioactive than the original form
Answer:
Option d: C₈H₉NO₂ = acetaminophen, analgesic
Explanation:
% composition of compound is:
63.57 g of C
6 g of H
9.267 g of N
21.17 g of O
First of all we divide each by the molar mass of the element
63.57 g / 12 gmol = 5.29 mol of C
6 g of H / 1 g/mol = 6 mol H
9.267 g of N / 14 g/mol = 0.662 mol of N
21.17 g of O / 16 g/mol = 1.32 mol of O
We divide each by the lowest value, in this case 0.662
5.29 / 0.662 = 8
6 / 0.662 = 9
0.662 / 0.662 = 1
1.32 / 0.662 = 2
Molecular formula of the compound is C₈H₉NO₂
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<em>Acetic acid, HC2H3O2</em>
First, calculate for the molar mass of acetic acid as shown below.
M = 1 + 2(12) + 3(1) + 2(16) = 60 g
Then, calculating for the percentages of each element.
<em> Hydrogen:</em>
P1 = ((4)(1)/60)(100%) = <em>6.67%</em>
<em> Carbon:</em>
P2 = ((2)(12)/60)(100%) = <em>40%</em>
<em>Oxygen</em>
P3 =((2)(16) / 60)(100%) = <em>53.33%</em>
<em>Glucose, C6H12O6</em>
The molar mass of glucose is as calculated below,
6(12) + 12(1) + 6(16) = 180
The percentages of the elements are as follow,
<em> Hydrogen:</em>
P1 = (12/180)(100%) = <em>6.67%</em>
<em>Carbon:</em>
P2 = ((6)(12) / 180)(100%) = <em>40%</em>
<em>Oxygen:</em>
P3 = ((6)(16) / 180)(100%) = <em>53.33%</em>
b. Since the empirical formula of the given substances are just the same and can be written as CH2O then, the percentages of each element composing them will just be equal.
