Calculate the theoretical yield (in grams) of magnesium oxide in your experiment in both trial 1 and 2 nsertterPage LayoutTimes New Rom 12Á ABIU abe x, x' Ab AReferences. MailingsFontAaSLab 5 Workbook (Empirical Formula) - Microsoft Word (Product Activation Failed)0.1963g-.1223g 0.074gTrial 14 Words: 446O Type here to searchReviewViewE-1E-5 * 21 ¶3. Using the law of conservation of mass, calculate the mass of oxygen in the product.Trial 1Trial 20.9g-0.564g-0.336gParagraph4. What is the experimental percent composition of magnesium and oxygen in the product of yourreaction?0.1223/0.1963*100=62.30% Magnesium0.074/0.1963*100= 37.69% OxygenAminte-Ad AaBb AaBbC AaBbC AaBbC AaBbCHeading 1Heading 2Heading 3Heading 4Heading 5Heading 9Styles5. Calculate the moles of magnesium used in this experiment.Trial 1Atomic mass of magnesium is 24.305 so 0.1223g/24.305-0.005 moleO ETrial 2564/.900 *100=62.67% Magnesium0.0336/900 37.33% OxygenTrial 20.564/24.305=0.023 moleWAFUSCE140 Insert Page LayoutanterTimes New Rom 12 AA Ал IM)BIU abe x X A aby?AReferences. MailingsFontLab 5 Workbook (Empirical Formula) - Microsoft Word (Product Activation Failed)Review4 Words: 446View*# *# 21 ¶E3--Paragraph6. Calculate the moles of oxygen atoms used in this experiment.Trial 1Atomic mass of O=16g/mol 0.0740/16g=0.0046 moleType here to searchMole of mag/mol of Ox=0.005/0.0046-1.08 moleAd AaBb AaBbc AaBbC AaBbC AaBbCHeading 1 Heading 21Heading 3Heading 5Heading 91Heading 4Styles7. Calculate the ratio between the number of moles of magnesium and the number of moles of oxygen inthe product. What is the calculated empirical formula of magnesium oxide?Trial 1Trial 2Trial 20.336/16=0.021 mole0.023/0.021 1.09 mole8. What is the actual empirical formula of magnesium oxide?How does this compare to the formula found by your experimentation?MgO9. What is the actual percent composition of magnesium oxide? How does this compare to the percentcomposition found in your experiment?Magnesium is 60.19% and Oxygen 39.81%.It compares fairyl well because the numbers are not far off from eachother in the least.4SEAChangeStyles140% -

Answered: Calculate the theoretical yield (in…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Calculate the theoretical yield (in grams) of magnesium oxide in your experiment in both trial 1 and 2

nsert
ter
Page Layout
Times New Rom 12
Á A
BIU abe x, x' Ab A
References. Mailings
Font
Aa
S
Lab 5 Workbook (Empirical Formula) - Microsoft Word (Product Activation Failed)
0.1963g-.1223g 0.074g
Trial 1
4 Words: 446
O Type here to search
Review
View
E-1E-5 * 21 ¶
3. Using the law of conservation of mass, calculate the mass of oxygen in the product.
Trial 1
Trial 2
0.9g-0.564g-0.336g
Paragraph
4. What is the experimental percent composition of magnesium and oxygen in the product of your
reaction?
0.1223/0.1963*100=62.30% Magnesium
0.074/0.1963*100= 37.69% Oxygen
Amin
te-
Ad AaBb AaBbC AaBbC AaBbC AaBbC
Heading 1
Heading 2
Heading 3
Heading 4
Heading 5
Heading 9
Styles
5. Calculate the moles of magnesium used in this experiment.
Trial 1
Atomic mass of magnesium is 24.305 so 0.1223g/24.305-0.005 mole
O E
Trial 2
564/.900 *100=62.67% Magnesium
0.0336/900 37.33% Oxygen
Trial 2
0.564/24.305=0.023 mole
WA
FUSCE
140

Insert Page Layout
anter
Times New Rom 12 AA Ал IM)
BIU abe x X A aby?
A
References. Mailings
Font
Lab 5 Workbook (Empirical Formula) - Microsoft Word (Product Activation Failed)
Review
4 Words: 446
View
*# *# 21 ¶
E
3--
Paragraph
6. Calculate the moles of oxygen atoms used in this experiment.
Trial 1
Atomic mass of O=16g/mol 0.0740/16g=0.0046 mole
Type here to search
Mole of mag/mol of Ox=0.005/0.0046-1.08 mole
Ad AaBb AaBbc AaBbC AaBbC AaBbC
Heading 1 Heading 21
Heading 3
Heading 5
Heading 91
Heading 4
Styles
7. Calculate the ratio between the number of moles of magnesium and the number of moles of oxygen in
the product. What is the calculated empirical formula of magnesium oxide?
Trial 1
Trial 2
Trial 2
0.336/16=0.021 mole
0.023/0.021 1.09 mole
8. What is the actual empirical formula of magnesium oxide?
How does this compare to the formula found by your experimentation?
MgO
9. What is the actual percent composition of magnesium oxide? How does this compare to the percent
composition found in your experiment?
Magnesium is 60.19% and Oxygen 39.81%.
It compares fairyl well because the numbers are not far off from eachother in the least.
4
SE
A
Change
Styles
140% -

Expert Solution

Step 1 Theoritical yield calculation in trial 1

Given:

$M a s s o f M g = 0.1223 g M a s s o f O_{2} = 0.074 g$

$M o l a r m a s s o f M g = 24.305 g / m o l M o l a r m a s s o f M g O = 40.304 g / m o l$

We know that,

$2 M g + O_{2} \to 2 M g O$

Here, limiting reactant is Mg.

$2 m o l o f M g p r o d u c e s = 2 m o l e o f M g O$

Therefore,

$2 m o l \times 24.305 g / m o l = 2 m o l \times 40.304 g / m o l$

$0.1223 g o f M g w i l l p r o d u c e = \frac{2 m o l \times 24.305 g / m o l}{2 m o l \times 40.304 g / m o l} \times 0.1223 g$

Hence, on calculation

$Theoritical yield of MgO = 0.2028 g$

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