71 terms

This set is written for the 'learn' format, it can be used in other formats but it is most effective when set on 'learn'. Hope this can help you out on chem 101 test #2 (unit 4 and 5).

Cis-trans Isomers

has identical bonding and substituents, only differs in 3-dimenstional arrangement of the substituents in space

Alcohols

classified by the number and/or position of the hydroxyl group

Alkynes

Unsaturated hydrocarbons containing at least one carbon to carbon triple bond

Hydration

A reaction involving the addition of water

Combustion

this reaction produces CO2, H2O and heat

Covalent Bond

Inter-atomic relationship created by the sharing of at least one pair of electrons

Carbon

Must follow the octet rule in order to achieve a stable configuration by sharing four electrons

Unsaturated Hydrocarbon

contain carbon to carbon double or triple bonds

Hydrogen and Halogen

shares one electron with carbon

Aliphatic hydrocarbons

hydrocarbons that contain only Hydrogen and Carbon

Benzene Ring

a ring of six carbon atoms with alternating single and double carbon-carbon bonds

Tertiary Carbon

Carbons that are covalently bonded to three other carbons

Alkanes

Saturated hydrocarbons containing only carbon to carbon single bonds (derivatives of methane)

Monohydric

Alcohol that contains only one -OH group

Tertiary Alcohol

the -OH group is covalently bonded to a tertiary carbon

Esterification; produces ester + H2O

organic acid + alcohol (in the presence of an acid catalyst and heat)

One disulfide molecule

The oxidation of two thiol molecules yields?

Ether + H2O

The dehydration of excess alcohol yields what?

Alkenes

Unsaturated hydrocarbons containing at least one carbon to carbon double bond

Aromatics

Cyclic organic compounds characterized by the presence of at least one 6 carbon benzene ring

Trihydric

Alcohol that contains three -OH groups

Structural Isomerism

Compounds that have identical molecular formulas but different structures

Primary Carbon

Carbons that are covalently bonded to one other carbon

Substituted Hydrocarbon

in which one or more hydrogen atoms are replaced by another atom or group of atoms ( aka functional groups)

Oxygen and Sulfur

shares up to two electrons with carbon

Saturated Hydrocarbon

contains only carbon to carbon single bonds

Organic Chemistry

branch that deals with carbon compounds, including those with no relationship to life.

Structural Formula

type of organic chemistry formula that shows the actual bonding of atoms to each other

Hydrolysis

reverse of esterification, requires heat and acid (H+) to catalyze the reaction (yields an organic acid + alcohol)

Tertiary Alcohol

when oxidized there is no reaction under normal circumstances

Hydrogenation

a reaction involving the addition of hydrogen

Substitution Reaction

This type of reaction is prevalent in benzenes

Molarity

the number of moles of substance dissolved in one liter of solution (ex. M and mol/L)

Molarity

What is this formula used for?

g=(mol/L)(L)(GMW)

g=(mol/L)(L)(GMW)

Step 1: Identify the formula needed

mg = (mEq/L) (mGEW) (L)

Step 2: Find the mGEW

Ca- 1 x 23.0 = 23.0

Cl- 1 x 35.5 = 35.5

O- 3 x 16 = 48.0

(23.0) + (35.5) + (48.0) = (106.5) =GMW

TPIV- find valence for ClO3, which is [ -1 ] then find the absolute value which is ( 1 )

mGEW = 106.5 / 1 = 106.5 mg/mEq

Step 3: convert mL to L

9mL = 0.009 L

Step 4: Plug in for the knowns

mg = ( 21.0 mEq/L) (106.5 mg/mEq) (0.009 L)

mg = 20.1285

Round

20.1 mg

mg = (mEq/L) (mGEW) (L)

Step 2: Find the mGEW

Ca- 1 x 23.0 = 23.0

Cl- 1 x 35.5 = 35.5

O- 3 x 16 = 48.0

(23.0) + (35.5) + (48.0) = (106.5) =GMW

TPIV- find valence for ClO3, which is [ -1 ] then find the absolute value which is ( 1 )

mGEW = 106.5 / 1 = 106.5 mg/mEq

Step 3: convert mL to L

9mL = 0.009 L

Step 4: Plug in for the knowns

mg = ( 21.0 mEq/L) (106.5 mg/mEq) (0.009 L)

mg = 20.1285

Round

20.1 mg

How much (mg) NaClO3 is needed to prepare 9mL of a 21.0 mEq/L NaClO3 solution? Round to the tenths place.

Step 1: Identify the formula needed

g= (mol/L) (GMW) (L)

Step 2: Convert mL to L

400mL = 0.4L

Step 3: Find the GMW of Na3SO6

Na- 3 x 23.0 = 69.0

S- 1 x 32.1 = 32.1

O- 6 x 16.0 = 96.0

(69) + (32.1) + (96) = 197.1

GMW= 197.1

Step 4: plug in for the knowns

g= (3.00 mol/Eq) (197.1) (0.4L)

236.5g

g= (mol/L) (GMW) (L)

Step 2: Convert mL to L

400mL = 0.4L

Step 3: Find the GMW of Na3SO6

Na- 3 x 23.0 = 69.0

S- 1 x 32.1 = 32.1

O- 6 x 16.0 = 96.0

(69) + (32.1) + (96) = 197.1

GMW= 197.1

Step 4: plug in for the knowns

g= (3.00 mol/Eq) (197.1) (0.4L)

236.5g

Determine how many grams of Na3SO6 is needed to prepare 400 mL of a 3.00 mol/L Na3SO6 solution. Round to the tenths place.

Alcohol

Identify the functional group in this formula

CH3CH2OH

CH3CH2OH

Thiol

Identify the functional group in this formula

CH3CH2SH

CH3CH2SH

Ether

Identify the functional group in this formula

CH3CH2OCH3

CH3CH2OCH3

n-butyl group

This formula represents an ethyl (alkyl) group CH2CH3, what group does this formula represent?

CH2CH2CH2CH3

CH2CH2CH2CH3

Prop (3)

Identify which prefix (root) should be used for a 3 carbon chain using IUPAC nomenclature rules.

Hept (7)

Identify which prefix (root) should be used for a 7 carbon chain using IUPAC nomenclature rules.

Secondary Alcohol

What does a ketone yeild in a reduction reaction?

Aldehyde

What type of group yields a primary alcohol upon reduction and an organic acid upon oxidation?

Sulfhydryl group

Some protein properties depend on the presence of this functional group and this group contributes to protein structure and activity?

*this one may be a little tricky

*this one may be a little tricky

Oxidation (OIL)

loss of hydrogen (or electron) and the gain of an oxygen

Halogenation

this reaction involves the substitution of Hydrogen with one or more Halogens and yields a halogenated alkane

Step 1: identify the formula needed

GEW=GMW/TPIV

Step 2: Find TPIV

common valence of (SO3) is [ - 2 ] multiply this by the subscript (3) and then take the absolute value of the answer

- 2 x 3 = - 6 = 6

TPIV= 6

Step 3: Find the GMW

multiply the number of atoms in each element by there atomic mass

Ca- 3 x 40.1 =120.3

S- 3 x 32.1 = 96.3

O- 9 x 16 = 144.0

(120.3) + (96.3) + (144.0) = 360.6 GMW

Step 4: plug in for the knowns and solve

GEW = 360.6 / 6 = 60.1 g/Eq

60.1 g/Eq

GEW=GMW/TPIV

Step 2: Find TPIV

common valence of (SO3) is [ - 2 ] multiply this by the subscript (3) and then take the absolute value of the answer

- 2 x 3 = - 6 = 6

TPIV= 6

Step 3: Find the GMW

multiply the number of atoms in each element by there atomic mass

Ca- 3 x 40.1 =120.3

S- 3 x 32.1 = 96.3

O- 9 x 16 = 144.0

(120.3) + (96.3) + (144.0) = 360.6 GMW

Step 4: plug in for the knowns and solve

GEW = 360.6 / 6 = 60.1 g/Eq

60.1 g/Eq

Find the GEW for a Ca3(SO3)3 compound.

11.6 moles

How many moles of solute are there in 4.3 L of a 2.7 mol/L solution?

Step 1: Identify formula needed

#moles = grams/GMW *simplified formula

Step 2: Find GMW of NaSO2

Na- 1 x23.0 = 23.0

S- 1 x 32.1 = 32.1

O- 2 x 16 = 32.0

(23.0) + (32.1) + (32.0) = 87.1g/mol = GMW

Step 3: plug in for the knowns and solve

#moles = 293.0 / 87.1 = 3.3639494483

round = 3.3

3.3 moles

#moles = grams/GMW *simplified formula

Step 2: Find GMW of NaSO2

Na- 1 x23.0 = 23.0

S- 1 x 32.1 = 32.1

O- 2 x 16 = 32.0

(23.0) + (32.1) + (32.0) = 87.1g/mol = GMW

Step 3: plug in for the knowns and solve

#moles = 293.0 / 87.1 = 3.3639494483

round = 3.3

3.3 moles

Determine the number of moles contained in a 293.0 grams of NaSO2.

Step 1: Identify the formula

g=(Eg/L)(GEW)(L)

Step 2: Find the GEW of NaCl4

First find the GMW

Na- 1 x 23.0 = 23.0

Cl- 4 x 35.5 = 142.0

(23.0) + (142.0) = 165.0

GMW = 165.0 g/Eg

Find the TPIV of NaCl4

(Cl) common valence of [ - 1 ]

(-1)x subscript (4) = -(4) take the absolute value of (-4), which will end up being (4)

TPIV = 4

GEW = 165.0 / 4= 41.25g/Eq

Step 3: covert mL into L

275mL = 0.275 L

Step 4: Plug in for the knowns and solve

(89.5g) = (Eq/L)(41.25g/Eq)(0.275L)

(89.5g) = (Eq/L)(11.34375)

(89.5) / (11.34375) = (Eg/L)(11.34375) / (11.34375)

Eg/L = 7.889807162534435

ROUND

7.9 Eq/L

g=(Eg/L)(GEW)(L)

Step 2: Find the GEW of NaCl4

First find the GMW

Na- 1 x 23.0 = 23.0

Cl- 4 x 35.5 = 142.0

(23.0) + (142.0) = 165.0

GMW = 165.0 g/Eg

Find the TPIV of NaCl4

(Cl) common valence of [ - 1 ]

(-1)x subscript (4) = -(4) take the absolute value of (-4), which will end up being (4)

TPIV = 4

GEW = 165.0 / 4= 41.25g/Eq

Step 3: covert mL into L

275mL = 0.275 L

Step 4: Plug in for the knowns and solve

(89.5g) = (Eq/L)(41.25g/Eq)(0.275L)

(89.5g) = (Eq/L)(11.34375)

(89.5) / (11.34375) = (Eg/L)(11.34375) / (11.34375)

Eg/L = 7.889807162534435

ROUND

7.9 Eq/L

Determine the Eq/L concentration of an NaCl4 solution that was prepared by adding 89.5 grams NaCl4 to a 275mL flask. Round to the tenths place.

Dehydration

this reaction type of reaction yields alkene + water in the presence of an acid

Organic acid + alcohol

Name the product of this reaction ESTERIFICATION: ester + water------- (requires heat and an acid catalyst)

Dihydric

Alcohols that contain two -OH groups are classified as?

Carboxylic Acid

What functional group is this?

O=C - OH

O=C - OH

Amide

Identify this functional group.

O=C - NH2

(product formed between a carboxylic acid and an amine)

O=C - NH2

(product formed between a carboxylic acid and an amine)

Molecular Formula

A formula used in organic chemistry that states the actual number of each kind of atom found in a molecule

Condensed Formula

Identify this type of organic chemistry formula:

CH3(CH2)2CH3

CH3(CH2)2CH3

Secondary Carbon

Carbons that are covalently bonded to two other carbons are classified as what type of carbon?

Nitrogen

This element shares up to three electrons with carbon

Molecular Formula

What type of formula is this?

C4H9O2

C4H9O2

Trans-Isomer

Structures with substituents on opposite sides of the carbon ring are called? (across the chain)

Ester

What type of functional group is this?

O=C - O

(compounds formed by the reaction between carboxylic acid and an alcohol)

O=C - O

(compounds formed by the reaction between carboxylic acid and an alcohol)

Amine

Identify the functional group in this condensed formula

CH3NH2

CH3NH2

CH2OHCHOHCH2OH

What is the condensed formula for 1,2,3 - Propanetriol?

Step 1: Identify the formula needed

g=(Eq/L)(GEW)(L)

Step 2: Find the GEW of Na(SO4)2

GMW

Na- 1 x 23.0 = 23.0

S- 2 x 32.1 = 64.2

O- 8 x 16 = 128

(23) + (64.2) + (128) = 215.2 g/mol

GMW = 215.2

TPIV

identify the common valence for (SO4), which is [ - 2 ]

Multiply this valence times the subscript (2)

(-2) x (2) = (- 4) take the absolute value of this number, which is (4)

TPIV=4

GEW=GMW/TPIV

GEW=215.2 / 4 = 53.8 g/Eq

Step 2: plug into the formula for the knowns and solve

(64.0g)=(1.9 Eq/L)(53.8 g/Eq)(L)

(64)=(102.22)(L)

(64) / (102.22) = (L) / (102.22)

L = 0.626100567

Step 3: convert L into mL

0.626100567L = 626.100567mL

Round to the nearest tenth

626.1mL

g=(Eq/L)(GEW)(L)

Step 2: Find the GEW of Na(SO4)2

GMW

Na- 1 x 23.0 = 23.0

S- 2 x 32.1 = 64.2

O- 8 x 16 = 128

(23) + (64.2) + (128) = 215.2 g/mol

GMW = 215.2

TPIV

identify the common valence for (SO4), which is [ - 2 ]

Multiply this valence times the subscript (2)

(-2) x (2) = (- 4) take the absolute value of this number, which is (4)

TPIV=4

GEW=GMW/TPIV

GEW=215.2 / 4 = 53.8 g/Eq

Step 2: plug into the formula for the knowns and solve

(64.0g)=(1.9 Eq/L)(53.8 g/Eq)(L)

(64)=(102.22)(L)

(64) / (102.22) = (L) / (102.22)

L = 0.626100567

Step 3: convert L into mL

0.626100567L = 626.100567mL

Round to the nearest tenth

626.1mL

What volume of solution (mL) an be prepared from 64.0 grams of Na(SO4)2 if the desired concentration is 1.9 Eq/L? Round to the nearest tenth.

Step 1: Identify the formula needed

g=(mol/L)(L)(GMW)

Step 2: Calculate GMW for CaCO3

Ca- 1 x 40.1 = 40.1

C- 1 x 12 = 12.0

O- 3 x 16 = 48.0

add (40.1) + (12.0) + (48.0) = 100.1 g/mol

GMW= 100.1

Step 3: Plug in the knowns and solve

(34.0g)=(2.7 mol/L)(L)(100.1g/mol)

(34.0)=(270.27)(L)

(34) / (270.27) = (270.27)(L) / (270.27)

(L)= (34) / (270.27)

L= 0.125800126

Round to the tenths place

0.1 L

g=(mol/L)(L)(GMW)

Step 2: Calculate GMW for CaCO3

Ca- 1 x 40.1 = 40.1

C- 1 x 12 = 12.0

O- 3 x 16 = 48.0

add (40.1) + (12.0) + (48.0) = 100.1 g/mol

GMW= 100.1

Step 3: Plug in the knowns and solve

(34.0g)=(2.7 mol/L)(L)(100.1g/mol)

(34.0)=(270.27)(L)

(34) / (270.27) = (270.27)(L) / (270.27)

(L)= (34) / (270.27)

L= 0.125800126

Round to the tenths place

0.1 L

What volume (L) of CaCO3 solution can be prepared from 34.0 grams of CaCO3 if a 2.7 mol/L CaCO4 solution is desired? Round to the nearest tenths place

Step 1: identify the formula needed

g = (mol/L)(GMW)(L)

Step 2: Calculate the GMW (including H2O)

Cu- 1 x 63.5 = 63.5

S- 1 x 32.1 = 32.1

O- 4 x 16 = 64.0

H2O- 6 x 18 = 108.0

add (63.5) + (32.1) + (64.0) + (108.0) = 267.6

GMW = 267.6 g/mol

Step 2: covert mL into L

275mL = 0.275 L

Step 3: Plug values into formula and solve

(g)= (3.2 mol/L)(267.6 g/mol)(0.275L)

g= 235.488

round to the tenths place

235.4g

g = (mol/L)(GMW)(L)

Step 2: Calculate the GMW (including H2O)

Cu- 1 x 63.5 = 63.5

S- 1 x 32.1 = 32.1

O- 4 x 16 = 64.0

H2O- 6 x 18 = 108.0

add (63.5) + (32.1) + (64.0) + (108.0) = 267.6

GMW = 267.6 g/mol

Step 2: covert mL into L

275mL = 0.275 L

Step 3: Plug values into formula and solve

(g)= (3.2 mol/L)(267.6 g/mol)(0.275L)

g= 235.488

round to the tenths place

235.4g

How much CuSO4 * 6HOH is required to prepare 275 mL of a 3.2 mol/L CuSO4 solution? Round to the tenths place

Step 1: Identify the formula needed

g=(Eg/L)(GEW)(L)

Step 2: Find the GEW (include H2O)

GMW

Cu- 1 x 63.5 = 63.5

S- 1 x 32.1= 32.1

O- 4 x 16 = 64.0

H2O- 3 x 18 = 54.0

add (63.5) + (32.1) + (64.0) + (54.0) = 213.6

GMW= 213.6 g/mol

TPIV

find the common valence of (SO4) which is [ - 2 ]

taking the absolute value of this number will be (2)

TPIV= 2

GEW= (213.6) / (2) = 106.8 g/Eq

Step 3: Convert mL into L

600mL= 0.6L

Step 4: Place values into the formula and solve

g=(1.2)(106.8)(0.6)

g=76.896

round to the tenths place

76.9g

g=(Eg/L)(GEW)(L)

Step 2: Find the GEW (include H2O)

GMW

Cu- 1 x 63.5 = 63.5

S- 1 x 32.1= 32.1

O- 4 x 16 = 64.0

H2O- 3 x 18 = 54.0

add (63.5) + (32.1) + (64.0) + (54.0) = 213.6

GMW= 213.6 g/mol

TPIV

find the common valence of (SO4) which is [ - 2 ]

taking the absolute value of this number will be (2)

TPIV= 2

GEW= (213.6) / (2) = 106.8 g/Eq

Step 3: Convert mL into L

600mL= 0.6L

Step 4: Place values into the formula and solve

g=(1.2)(106.8)(0.6)

g=76.896

round to the tenths place

76.9g

Determine how much CuSO4*3H2O is required to prepare 600 mL of a 1.2 Eq/L CuSO4 solution. Round to the tenths place.

Aldehydes

What functional group is this?

O=C - H

O=C - H

Ketone

Identify the functional group in this condensed formula

CH3CH2CH=O

CH3CH2CH=O