0.6 sample an unknown organic acid found in muscle cells is burned in air and found to contain 0.24g of carbon, 0.04g of hydrogen with the rest being oxygen. If the molecular weight of the substance is 90g/mol what is the emperical and molecular formula

Molecular formula is, C₃H₆O₃

Empirical formula (with the lowest subscript) → CH₂O

Explanation:

We assume that the organic acid's mass is 0.6 g. We know that, in 0.6 g of compound we have 0.24 g of C and 0.04 g of H, then, we have

(0.6 - 0.24 - 0.04) = 0.32 g of O.

We determine each mol:

0.24 g /12 g/mol = 0.02 mol of C

0.04 g/ 1g/mol = 0.04 mol of H

0.32 g/ 16 g/mol = 0.02 mol of O

1 mol of acid weighs 90 g/mol.

In 0.6 g of acid, we have, (0.6 g / 90g/mol) = 0.0067 moles

Let's find out the formula with rules of three:

0.00667 mol of acid have 0.02 moles of C

1 mol of acid may have (0.02 /0.00667) = 3 mol of C

0.00667 mol of acid have 0.04 moles of H

1 mol of acid may have (0.04 /0.00667) =  6 mol of H

0.00667 mol of acid have 0.02 moles of O

1 mol of acid may have (0.02 /0.00667) = 3 mol of O

Molecular formula is, C₃H₆O₃

We confirm by the molar mass: 12g/mol . 3 + 6 . 1g/mol + 32g/mol . 3 = 90

Empirical formula (with the lowest subscript) → CH₂O

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.

3 1/4 %

explanation: