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Gr.9 po ito pag nasagot nyopo automatic na brainliest po kayoo
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PART II. Calculating the Empirical and Molecular Formula of an Unknown Compound
1. Suppose a chemical analysis shows that a complex carbohydrate contains 40% C, 6.71% Hand 53.3%O by mass. To get the empirical formula of the unknown carbohydrate, change first the percentages into decimal. a. How many grams of each element are there?
Carbon =________
Hydrogen = _______
Oxygen = ________
2. Calculate the number of moles of each element by dividing their mass with their atomic mass. a. How many moles of each element are there? Carbon =________ Hydrogen =______Oxygen =________
b. What is the smallest number of moles that you computed among the three elements? ___________
3. Divide the moles of each element by the smallest number of mole. A. What are the resulting mole number ratio per element? Carbon =_________
Hydrogen =________
Oxygen=__________
4. Make the mole number ratio the subscript for each element. If the subscript is 1, it is understandable to not write the subscript anymore.
A. What is the resulting empirical formula?____________________
5. After knowing the empirical formula of the unknown carbohydrate, you can now calculate its molecular formula given that its molar mass is approximately 870 g/mol.
A. What is the molar mass of the empirical formula?__________
B.After dividing the molar mass of the unknown carbohydrate to the molar mass of the empirical formula, what is the resulting multiplier?______________________
C. What is the molecular formula of the unknown carbohydrate?_________​


Gr9 Po Ito Pag Nasagot Nyopo Automatic Na Brainliest Po Kayoo Ps Mahirap Po Yan Itype Isa Isa Sana May Matino Po Na MagsagotPART II Calculating The Empirical An class=

Sagot :

Answer:

Answer:

Carbon = 40g

Hydrogen = 6.71g

Oxygen = 53.3g

Carbon = 3.330g/mol

Hydrogen= 6.644g/mol

Oxygen = 3.331g/mol

B. 3.330g/mol

Carbon= 1

Hydrogen = 2

Oxygen = 1

A. CH2O

A. 100.01 g/mol

B. 8.7

C. C9H18O9

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