Acompound is 54.53% c, 9.15% h, and 36.32% o. what is its empirical formula?

the molar mass of the compound is 132 amu. what is its molecular formula?

The empirical formula : C₂H₄O

The molecular formula : C₆H₁₂O₃

The empirical formula is the smallest comparison of atoms of compound forming elements.

A molecular formula is a formula that shows the number of atomic elements that make up a compound.

(empirical formula) n = molecular formula

The principle of determining empirical formula and molecular formula

Percentage of elements A and B in AxBy compounds:

% A in AₓBy = (x x Ar A / Mr AₓBy) x% AₓBy

% B in AₓBy = (y x Ar B / Mr AₓBy) x% AₓBy

It is known about the% mass of each element

To find the empirical formula, we look for its mole ratio (% mass/ its relative atomic mass)

C = 54.53 / 12 = 4.544

H = 9.15 / 1 = 9.15

O = 36.32 / 16 = 2.27

To simplify the comparison as an integer, we divide by the mole ratio of the smallest element

C = 4.544 / 2.27 = 2

H = 9.15 / 2.27 = 4

O = 2.27/2.27 = 1

So the empirical formula:

C₂H₄O

We look for the molecular formula to get the number of atoms making up the compound

(C₂H₄O) n = molecular formula

(2.12 + 4.1 + 16) n = 132

(44) n = 132

n = 3

So molecular formula = C₆H₁₂O₃

the molar mass of a substance

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Determine the empirical and molecular formulas

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Determine the molecular formula of the compound

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Keywords: empirical formula, molecular formula, mass, mole

imma come back to you hold on

1. the question asks the most appropriate, which seems to rule out "all of the above." your choice appears to be the correct one.

2. the protocol described is good for identifying correlation, but not cause and effect. for cause/effect, you need to devise a protocol that will assign the treatment (or not) to groups that have essentially the same "demographics" with respect to all the other variables. your choice regarding control group seems appropriate. random assignment of treatment (choice d) may be appropriate also. (i think you can argue that b and d are just different descriptions of the same method.)

3. d will randomize the birds, but may not guarantee there are 10 in each group. e will do both, so seems the better choice.

comment on question 2

based on the problem statement, it seems clear that confident, attentive drivers are more likely to use daytime running lights. so, the protocol might need to be devised in such a way that the lights would be turned on or off without the driver's knowledge. (an attentive driver will figure it out.) assigning a confident, attentive driver to a group required to drive without running lights might mess with his/her confidence and attentiveness. you don't want your experiment to do that. (nor do you want the experiment to subject a driver to greater risk of accident as a consequence of driving without running lights.) experiments involving human safety are always problematic.