Thus, the definition of a mole is as follows.
One mole of a substance expressed many mengan-dung number of particles equal to the number partikeldalam 12.0 grams of C-12 isotope.
For example:
1 1 mole of Na element contains 6.02 x 1023 atoms of Na.
2 1 mole of water contains 6.02 x 1023 molecules of water.
3 1 mol NaCl ionic compound contains 6.02 x 1023 ions Na + and Cl-ions 6.02 x 1023.
Mol relationship with the number of particles
Relationship with the number of moles of particles can be formulated:
quantity (in moles) = number of particles / NA
or
the number of particles = moles x NA
Example question:
A sample containing 1.505 x 1023 molecules of Cl2, how many moles of the Cl2 content?
Answer:
The quantity (in moles) Cl2 Cl2 = number of particles / NA
= 1.505 x 1023 / 6.02 x 1023
= 0.25 mol
Mol relationship with Mass
Before discussing the relationship of moles to mass, you need to remember in advance of the Relative Atomic Mass (Ar) and relative molecular mass (Mr). Still remember? Then we check your memory to do the problems below.
Calculate Mr. H2SO4 (Ar H = 1, S = 32, and O = 16)!
Known relative atomic mass (Ar) some elements as follows.
Ca = 40
O = 16
H = 1
Determine the relative molecular mass (Mr) compound Ca (OH) 2!
Already remember? So we went straight to the next matter is the molar mass.
Molar mass states that mass is owned by 1 mole of a substance, which is the same as Ar or Mr
For elements:
1 mole of an element = Ar grams, it can be formulated:
= Mass of 1 mole of a substance expressed in grams of Ar
or
The molar mass of the substance is a substance Ar = great g / mol
To compound:
1 mole of the compound = Mr. grams, it can be formulated:
1mol mass of substance = Mr substance expressed in grams
or
The molar mass of the substance of Mr. substance = grams / mole
So the difference between molar mass and relative molecular mass is in the unit. Molar mass has units of g / mol while the relative molecular mass has no units.
The relationship between the mole mass is:
Quantity (in mol) = mass of the compound or element (g) / molar mass compounds or elements (g / mol)
Mol relationship with Volume
a. Gas in the standard state
Measurement of the quantity of gas depends on the temperature and pressure of the gas. If the gas is measured at the standard state, then the volume is called the molar volume. The molar volume is the volume of 1 mole of gas measured at standard state. Standard state is the state at a temperature of 0 ° C (or 273 K) and a pressure of 1 atmosphere (or 76 cmHg or 760 mmHg) or abbreviated STP (Standard Temperature and Pressure).
The magnitude of the molar volume of the gas can be determined by the ideal gas equation: PV = nRT
P = pressure = 1 atm
n = moles = 1 mole of gas
T = temperature in Kelvin = 273 K
R = gas constant = 0.082 liter atm / mol K
then:
P V = nRT
V = 1 x 0,082 x 273
V = 22,389
V = 22.4 liters
Thus, a standard volume = VSTP = 22.4 L / mol.
Can be formulated: V = n x Vm
n = number of moles
Vm = VSTP = molar volume
Example question:
1) What is the quantity (in moles) of hydrogen gas whose volume is 6.72 liters, when measured at a temperature of 0 ° C and a pressure of 1 atm?
Answer:
The quantity (in moles) H2 = volume of H2 / VSTP
= 6.72 L / 22.4 mol / L
= 0.3 mol
2) Calculate the mass of 4.48 liters of C2H2 gas measured at standard state!
Answer:
Quantity (in mol) of C2H2 = C2H2 volume / VSTP
= 4.48 / 22, 4
= 0.2 mol
C2H2 mass = moles x molar mass of C2H2
= 0.2 mol x 26 g / mol
= 5.2 grams
3) Calculate the volume of 3.01 x 1023 molecules of NO2 were measured at a temperature of 0 ° C and a pressure of 76 cmHg!
Answer:
quantity (in moles) NO2 = number of particles / NA
= 3.01 x 1023 particles / 6.02 x 1023 particles / mol
= 0.5 mol
NO2 volume = moles x VSTP
= 0.5 moles x 22.4 L / mol
= 11.2 liters
b. Gas in the state of nonstandard
If the volume of gas measured at the state of the ATP (Am-bient Temperature and Pressure) or better known as the state of the non-STP then using the formula:
P V = n R T
P = pressure, force P is the atmosphere (atm)
V = volume, V is the unit liters
n = moles, n is the mole unit
R = gas constant = 0.082 liter atm / mol K
T = temperature, unit T is Kelvin (K)
Example question:
Determine the volume of 1.7 grams of ammonia gas were measured at a temperature of 27 ° C and a pressure of 76 cmHg!
Answer:
n = mass of ammonia / ammonia molar mass
= 1.7 g / 17 g / mol
= 0.1 mol
P = (76 cmHg / 76 cmHg) x 1 atm = 1 atm
T = (t + 273) K = 27 + 273 = 300 K
P V = n R T
1 atm × V = 0.1 × 0.082 mol L atm / mol K × 300 K
V = 2.46 L
Relationship with the mass mole, Avogadro's number and volume can be summarized in the chart below.
In the previous material has been described that the ratio of the coefficient states comparison of the number of particles and the volume ratio, while the mole acts as a particle divided by Avogadro's number. Comparison of the ratio of the coefficient states of particles, then the ratio of the coefficient is also a mole ratio.
Thus, it can be concluded that:
Comparison coefficient = ratio of the volume
= Ratio of the number of particles
= Mole ratio
For example, the reaction: N2 (g) + 3 H2 (g) → NH3 (g)
a. Comparison of the volume of N2 (g): H2 (g: NH3 (g) = 1: 3: 2
b. Comparison of the number of particles N 2 (g): H2 (g): NH3 (g) = 1: 3: 2
c. Comparison mol N2 (g): H2 (g): NH3 (g) = 1: 3: 2
example Problem
a. In reaction to the formation of ammonia gas (NH3) from nitrogen and hydrogen gas, nitrogen gas is reacted if it was 6 mol, then specify:
1) the number of moles of hydrogen gas are needed;
2) the number of moles of ammonia gas produced!
Answer:
1) N2 (g) + 3 H2 (g) → 2 NH3 (g)
Mol H2 = (coefficient of H2 / N2 coefficient) x moles of N2
= (3/1) x 6 = 18 mol
2) mol NH3 = (coefficient of NH3 / N2 coefficient) x moles of N2
= (2/1) x 6 = 12 mol
limiting reagent
If a box is available in the 6 nuts and 10 bolts, then we can make 6 pairs of nut-bolts. Bolt the remaining 4 pieces, while the nuts are exhausted. In a chemical reaction, if the mole ratio of the reactant substances are not the same as the ratio of the coefficient, then there reactants are depleted first. Reactant is called the limiting reagent.
Example question:
In the reaction of 0.5 moles of N2 gas with 2.5 mol H2 gas according to the reaction equation:
N2 (g) + 3 H2 (g) → 2 NH3 (g)
specify:
a. the limiting reagent;
b. how many grams of substance is left?
(Ar N = 14 and H = 1)!
Answer:
Finding a residual reactant mole and the completely reacted
N2 (g) + 3 H2 (g)
At first, 0.5 mol 2.5 mol
Which react: 0.5 mol 1.5 mol
After reaction: 0 mol 1.0 mol
Reagent is left over as much as 1.0 mol H2
H2 is the rest mass = moles x Mr residual
= 1.0 × 2
= 2 grams
levels of Substance
At the time of heat illness sister, your mother told mem-buy alcohol 70% at the pharmacy. Do you know what that means 70% alcohol? That is in 100 mL of solution containing 70 mL of alcohol and 30 mL of water. Similarly, if you buy a packaged food product containing vitamin C 1%. That is in 100 grams of food containing 1 gram of vitamin C content of substances commonly expressed in percent by mass (% mass). To get the mass percent using the formula:
% X in substance = (X mass / mass of substance) x 100%
Example question:
Calculate the mass of caffeine in a cup of coffee (200 grams) whose levels are 0.015%!
Answer:
% Mass of caffeine = (mass of caffeine / coffee mass) x 100%
0.015% = (mass of caffeine / 200) x 100%
Mass caffeine = 0.03%
2 Determine the percent C in glucose (C6H12O6), if known Ar C = 12, O = 16, and H = 1!
Answer:
Mass% C = ((number of Ar atoms C x C) / Mr. glucose) / 100%
= ((6 x 12) / 180) / 100%
Empirical formula and Molecular formula
The chemical formula is divided into two, namely the empirical formula and molecular formula. The empirical formula is a chemical formula that describes the smallest mole ratio of the atoms making up the compound.
One way to determine the empirical formula and molecular formula can be done the following steps.
Mass percent of each element → → mole mole ratio of the elements of data → Mr → empirical formula → molecular formula.
The molecular formula is the actual formula of a compound. Molecular formula can be determined if the relative molecular mass is known. The following example problems is one way to determine the empirical formula and molecular formula.
Example question:
A technician chemical burn 4.5 grams of sample Senya-wa-containing organic C, H, and O. When used pure oxygen gas, results in 6.6 grams 2.7 grams H2O CO2dan. specify:
1 The empirical formula of the organic compound (ARC = 12, O = 16, and H = 1);
2 molecular formula of the organic compound if its known Mr = 30!
Answer:
C in the CO2 mass
= ((Number of Ar atoms C x C) / Mr. CO 2) x mass of CO2
= ((1 x 12) / 44) x 6.6 grams = 1.8 grams
Quantity (in mol) C
C = mass / Ar C
= 1.8 / 12 = 0.15 mol
Mass of H in H2O
= ((Number of H x Ar H) / Mr H2O) x mass H2O
= ((2 x 1) / 18) x 2.7 gram
= 0.3 grams
Quantity (in mol) of H
= Mass of H / Ar H
= 0.3 / 1 = 0.3 mol
O = mass of the sample mass - mass of C - H mass
= 4,5 - 1.8 to 0.3 = 2.4 grams
Quantity (in mol) of O
= Mass of O / Ar O
= 2.4 / 16 = 0.15 mol
Comparison mol C: mol H: mol O = 0.15: 0.3: 0.15
= 1: 2: 1
Thus, the carbon compounds empirical formula is CH2O.
= Empirical formula (CH2O) n
then: Mr = (CH 2 O) n
30 = (12 + (2 x 1) + 16) n
30 = 30N
n = 1
Thus, the molecular formula of carbon compounds is (CH 2 O) 1 = CH 2 O or formic acid.
salt hydrates
You would never hear a cast (CaSO4.2H2O) which is used to connect bone or salt english / epsom salt (MgSO4.7H2O) used for laxative. Both of these compounds are examples of salt hydrates. Hydrate salt is salt water binding. If the salt hydrate releases water-bound crystal called anhydrous salt. How to find the amount of water bound to the salt crystal hydrates is the formula:
x = mole H2O / mol salt hydrate
Example question:
A total of 8.6 grams of salt hydrate crystal is heated until all the water evaporates and form 6.8 grams of CaSO4. If the statue = 40, O = 16, S = 32, and H = 1, set
formula of the salt hydrate!
Answer:
Quantity CaSO4 CaSO4 = mass / Mr CaSO4
= 6.8 / 136 = 0.05 mol
Mass of water = mass of salt hydrate - mass of anhydrous salt
= 8.6 to 6.8 = 1.8 grams
The quantity of water = mass of water / Mr water
= 1.8 / 18 = 0.1 mol
x = mole H2O / mol CaSO4
= 0.1 / 0.05
= 2
So, the formula is CaSO4 salt hydrates. 2H2O
May be useful ^ _ ^
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