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Problems

Hey guys. Some hijo de puta deleted my post, but I just need a SECOND opinion on some work I've done. Some of the immaturity by certain moderators is astounding. :rolleyes:

My answers are in italic.

1)

State the oxidation state, coordination number and d^n configuration for [PtCl2(NH3)2].

I'm confident about the first two parts (+2 and 6 respectively), but I'm not sure about the d^n configuration as f orbitals are involved. My answer is 8, but I just need clarification.

Reply 1

TOTB

2)

Ethyl caprylate is an ester that is sometimes called "cognac essence". Combustion analysis shows to to be 71.89% C, 12.13% H and 15.98% O by mass. Hydrolysis of the ester yields ethanol and an acid. The RMM of the acid is 172. What is the formula of ethyl caprylate?

C: 71.89/12 = 5.99
H: 12.13/1 = 12.13
O: 15.98/16 = 1.00

Ratio of C :h:

:O = 48:97:8

C48H97O

Surely there's something wrong here?

Reply 2

TOTB

3)

a) Write a balanced equation for the formation of methoxymethane (CH3OCH3) from its elements.

b) Calculate the standard enthalpy of formation of ethoxyethane given that:
deltaHc (C,s) = -393.5kJ per mole
deltaHc (H2,g) = -286kJ per mole
deltaHc (CH3OCH3) = -1455kJ per mole

a) 2C + 3H2 + 0.5O2 -> CH3OCH3

b) This is where I'm stuck. Surely, I have to draw a cycle of some sort. Maybe I'm overcomplicating a few things.

Reply 3

TOTB

4)

Suggest organic products for the reaction of but-2-yne with excess Br2.

I'm thinking of but-2-ene, and then butane?

Reply 4

TOTB

5)

Compound X contains only carbon, hydrogen, nitrogen and oxygen. Combustion of 0.157g of X produced 0.213g of CO2 and 0.0310g of H2O. In another experiment, 0.103g of X produced 0.023g of NH3. Determine the empirical formula of X.

0.213g of CO2 contains 0.00357mol of C and 0.00714mol of O
0.0310g of H2O contains 0.00172mol of O and 0.00344mol of H

0.023g of NH3 contains 0.00135mol of N and 0.00406mol of H
Equate this from 0.103g of X to 0.157g of X:
We have 0.00206mol of N and 0.00619mol of H

Total moles of C = 0.00357mol
H = 0.00963mol
O = 0.00886mol
N = 0.00206mol

Ratio of C :h:

:N:O

1.73 : 4.67 : 4.30 : 1.00

From here, I find it very difficult to produce a good empirical formula.

Reply 5

TOTB

Thanks in advance. :smile:

Reply 6

Flugtag

For 2, it is correct up to here:

C: 71.89/12 = 5.99
H: 12.13/1 = 12.13
O: 15.98/16 = 1.00

but now what you need to do is to divie each figure by the smallest number. so you would get in the end:
C6H12O
round them up/down of course

Reply 7

TOTB

markfung

Grr, so 12.13 is roughly equal to 12 then? I was trying to multiply by all sorts to get a whole number. :mad:

Thank you. :smile:

Reply 8

Flugtag

that is correct. Im not too sure onn the others so i wont do anything in case i get it all wrong!

Reply 9

TOTB

markfung

that is correct. Im not too sure onn the others so i wont do anything in case i get it all wrong!

Okay. No problem! :smile:

Anyone else?

Reply 10

TOTB

Bump.

Reply 11

charco

Study Forum Helper

¡hijo de puta es un poco fuerte, pero como me han quitado la firma de mi perfil estoy de acuerdo - son cabrones!

[PtCl2(NH3)2] - the Pt is in the oxidation state II therefore the d^n configuration is simply the element configuration less the two s electrons

b) Calculate the standard enthalpy of formation of ethoxyethane given that:
deltaHc (C,s) = -393.5kJ per mole
deltaHc (H2,g) = -286kJ per mole
deltaHc (CH3OCH3) = -1455kJ per mole

here you have to construct the equation required from the equations given...
equation 1 - C + O2 --> CO2
equation 2 - H2 + 1/2O2 --> H2O
equation 3 - CH3OCH3 + 3O2 --> 2CO2 + 3H2O

mtb 3 in equation 2 to get 3H2 + 3/2O2 --> 3H2O
mtb 2 in equation 1 to get 2C + 2O2 --> 2CO2
add these two equations together
3H2 + 2C + 7/2O2 --> 2CO2 + 3H2O
now subtract equation 3 from this answer and that gives you the equation for the formation of methoxymethane from its elements - do the same operations with the energy values and you have the answer

Suggest organic products for the reaction of but-2-yne with excess Br2.

why not just add bromine across the unsaturation? This would give 2,2,3,3, tetrabromobutane

Reply 12

charco

Study Forum Helper

TOTB

5) X contains only carbon, hydrogen, nitrogen and oxygen. Combustion of 0.157g of X produced 0.213g of CO2 and 0.0310g of H2O. In another experiment, 0.103g of X produced 0.023g of NH3. Determine the empirical formula of X.

you are going about it in the wrong way. You should first calculate the percentage of the elements in the compound remembering that the oxygen cannot be equated (it comes from the air on combustion) and must be obtained by subtraction of the other percentages.

%C = (12/44 x 0.213)/0.157 x 100 = 37.0%
%H = (2/18 x 0.031)/0.157 x 100 = 2.2%
%N = (0.023 x 14/17)/0.103 x 100 = 18.4%

these percentages add up to 57.6%
leaving 100 - 57.6 percentage oxygen = 42.4%

now divide by the elements mass
C = 37/12 = 3.08
H = 2.2/1 = 2.2
N = 18.4/14 = 1.314
O = 42.4/16 = 2.6625

divide through by smallest
C = 2.34
H = 1.67
N = 1
O = 2.0

mutiplying everything by 3 gives C7H5N3O6

Reply 13

TOTB

charco

¡hijo de puta es un poco fuerte, pero como me han quitado la firma de mi perfil estoy de acuerdo - son cabrones!

Gracias!

Reply 14

TOTB

charco

[PtCl2(NH3)2] - the Pt is in the oxidation state II therefore the d^n configuration is simply the element configuration less the two s electrons

Okay, Pt has an electronic configuration of [Xe] 6s2 4f14 5d8. So assuming that f electrons are ignored (?), then Pt2+ will have a d^n configuration of 8?