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Tanya [424]
3 years ago
13

The accepted value for the percent by mass of water in a hydrate is 36.0%. In a laboratory activity, a student determined the pe

rcent by mass of water in the hydrate to be 37.8%. What is the percent error for the student's measured value?
(1) 5.0% (3) 1.8%

(2) 4.8% (4) 0.05%
Chemistry
1 answer:
valentina_108 [34]3 years ago
4 0
The original or accepted value for the percent by mass of water in a hydrate = 36%
Percen by mass of water in the hydrate determined by the student
in the laboratory = 37.8%
So the difference between the actual and the percent by mass in water determined by student = (37.8 - 36.0)%
             = 1.8%
So the percentage of error made by the student = (1.8/36) * 100 percent
                                                                             = (18/360) * 100 percent
                                                                             = ( 1/20) * 100 percent
                                                                              = 5 percent
So the student makes an error of 5%. Option "1" is the correct option.
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To formulate the empirical formula, we need to follow some steps:

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Moles of Copper =\frac{\text{Given mass of Copper}}{\text{Molar mass of Copper}}=\frac{0.786g}{63.5g/mole}=0.0124moles

Moles of Chlorine = \frac{\text{Given mass of Chlorine}}{\text{Molar mass of Chlorine}}=\frac{1.328g}{35.5g/mole}=0.0374moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0124 moles.

For Copper = \frac{0.0124}{0.0124}=1

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