COUNTS CHEMICAL

COUNTS CHEMICAL

Count chemicals are ways of calculating oriented basic laws of chemistry.

In this case will be given a variety of sample questions and their chemical count pembahasanya.

Examples of questions:

1. What percentage levels of calcium (Ca) in the calcium carbonate? (Ar: C = 12; O = 16; Ca = 40)

Answer:

1 mol CaCO, containing 1 mol Ca + 1 + 3 mol mol C O

Mr CaCO3 = 40 + 12 + 48 = 100

So the level of calcium in CaCO3 = 40/100 x 100% = 40%

2. A total of 5.4 grams of aluminum metal (Ar = 27) was treated with excess dilute hydrochloric acid according to the reaction:

2 Al (s) + 6 HCl (aq) 2 AlCl 3 (aq) + 3 H2 (g)

How many grams of aluminum chloride, and how many liters of hydrogen gas generated at standard conditions?

Answer:

From equation can be expressed

2 moles of Al x 2 mol AlCl3  3 mol H2

5.4 grams of Al = 5.4 / 27 = 0.2 mol

So:

AlCl3 formed = 0.2 x Mr AlCl3 = 0.2 x 133.5 = 26.7 grams

H2 gas volume produced (0 ° C, 1 atm) = 3/2 x 0.2 x 22.4 = 6.72 liter

3. An iron ore containing 80% Fe2O3 (Ar: Fe = 56; O = 16). These oxides are reduced with CO gas to produce iron.

How many tons of iron ore required to make 224 tons of iron?

Answer:

1 mol Fe2O3 containing 2 mol Fe

then: Fe2O3 = mass (Mr Fe2O3 / 2 Ar Fe) mass x Fe = (160/112) x 224 = 320 tonnes

So the iron ore required = (100/80) x 320 tonnes = 400 tonnes

4. To determine the crystal waters of 24.95 grams of copper sulfate salt crystals are heated until all the water evaporates. After heating the mass of salt into 15.95 grams. How many crystalline water contained in the salt?

Answer:

suppose formula thereof is CuSO4. xH2O

CuSO4. xH2O  CuSO4 + xH2O

24.95 grams of CuSO4. xH2O = 159.5 + 18x mol

15.95 grams of CuSO4 = 159.5 mol = 0.1 mol

according to the equation above can be stated that:

the number of moles CuS04. xH2O = mol CuSO4; so that the equation

24.95 / (159.5 + 18x) = 0.1  x = 5

So the formula thereof is CuS04. 5H2O

Empirical formula and Molecular formula

The empirical formula is the simplest formula of a compound.

This formula simply states comparison of the number of atoms contained in the molecule.

The empirical formula of a compound can be determined if known one:

- Mass and Ar each element

-% Mass and Ar each element

- Mass ratio and Ar each element

Molecular formula: if the empirical formula is already known and Mr well known, molecular formula can be determined.

Example: A compound den C C H contains 6 grams and 1 gram H.

Determine the empirical formula and the molecular formula of the compound when known Mr her = 28!

Answer: C mol: mol H = 6/12: 1/1 = 1/2: 1 = 1: 2

So the empirical formula: (CH2) n

When Mr compounds were = 28, then: 12n + 2n = 28   14N = 28 n = 2

So the molecular formula: (CH2) 2 = C2H4

Example: For a 20 ml oxidize hydrocarbons (CxHy) in a gaseous state required oxygen as much as 100 ml and 60 ml CO2 generated. Determine the formula of the hydrocarbon molecules!

Answer: The equation combustion of hydrocarbons in general

CxHy (g) + (1/4 x + y) O2 (g) x CO2 (g) + 1/2 y H2O (l)

Reaction coefficient indicates mole ratio of the substances involved in the reaction.

According to Gay Lussac gases at p, t the same, the number of moles is directly proportional to the volume

Then:

CxHy mol: mol O2: mol CO2 = 1: (x + 1 / 4y): x

20: 100: 60 = 1: (x + 1 / 4y): x

1: 5: 3 = 1: (x + 1 / 4y): x

or:

1: 3 = 1: x x = 3

1: 5 = 1: (x + 1 / 4y) y = 8

So the formula of hydrocarbons are: C3H8