# data[].attributes.body
Lavani
a)The emprical formula is the chemical formula that shows the sinplest ratio of atom of element in compound.For example the emprical formula of ethene is CH2 and the molecular formula of ethene is (CH2)2=C2H4.
b)Mass of Mg
24.90-23.70=1.2g
c)Mass of O
25.70-24.90=0.8g
d)To remove oxide layer on its surface
e)Mol of Mg
0.05 mol
f)Mol of O
0.05
g)MgO
h)White fumes is produce when the magnesium ribbon start to burns.
i)To avoid the white fumes escape
j)To allow oxygen to enter

# data[].attributes.body
a) Empirical formula is the formula that show the simplest whole number ratio of atom of element in a compound. Molecular formula is the chemical formula that show the actual number of atoms of each element present in one molecule of the compound.
b)mass of magnesium=24.90g-23.70g=1.20g
c)mass of oxygen=25.70g-24.90g=0.80g
d)To remove the oxide layer of magnesium oxide.
e)1.2÷24=0.05mol
f)0.8÷16=0.05mol
g)MgO
h)White fumes produced when the magnesium ribbon burned.
i)To avoid white fumes from escaping to surrounding.
j)To allow oxygen gas enter the crucible to continue burning process.

# data[].attributes.body
a)Empirical formula of a compound shows the simplest ratio of atoms of the element that combine to form the compound.For example,the molecular formula of ethane is C2H6,simplest ratio of the element is C:H=1:3.So the empirical formula is CH3.The molecular formula of the compound shows the actual numbers of the atoms of the elements that combine to form the compound.For an example H2O.

b)24.90-23.70=1.2g
c)25.70-24.90=0.8g
d)To remove the oxide layer on the surface of the magnesium ribbon.
e)mole of magnesium=0.05
f)mole of oxygen=0.05
g)MgO
h)When the magnesium ribbon starts to burn,white fumes are produced.
i)to prevent the white fumes escape.
k)To make sure the oxygen gas to enter into the crucible for the combustion of the magnesium ribbon.

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Eamyra
a)The empirical formula of compound shows the simplest ratio of atom of elements to combine to form the compund.For example,the molecular formula of ethene is C2H4.The simplest ratio of the elements is C:H =1:2.so the empirical formula is CH2.the molecular formula of the compound shows the actual numbers of the atoms oh the elements that combine to form the compound.For example,C2H4.
b) Mass of Mg=1.2g
c) Mass of O=0.8g
d) To remove the oxide layer on its surface of magnesium ribbon.
e) Mole of Mg=0.05g
f)Mole of O=0.05g
g)MgO
h)White smoke produced when the magnesium oxide start burning.
i) To prevent the white fumes of magnesium oxide from escape.
k) To allow the oxygen enter the crucible to help magnesium in burning.

# data[].attributes.body
(a)Empirical formula of a compound shows the simplest ratio of the atoms,of the elements that combine to form the compound.For example,the molecular formula of ethane is C2H6 ,simplest ratio of the elements are C:H =1:3, then the empirical formula is CH3. The molecular formula of the compound shows the actual numbers of the atoms of the elements that combine to form the compound.For example, H2O.

(b)24.9-23.7=1.2g

(c)25.7-24.9=0.8g

(d)To remove the oxide layer on its surface.

(e)1.2÷24=0.05mol

(f)0.8÷16=0.05mol

(g)MgO

(h)white smoke is observed

(i)To prevent the white smoke from escaping

(k)To allow the oxygen from the air enter for the combustion of the magnesium.

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Exercise1: Empirical formulae of magnesium oxide

An experiment was conducted in a laboratory to determine the empirical formulae of magnesium oxide by heating magnesium strongly in a crucible. Magnesium ribbon was cleaned with a sand paper and it is coiled loosely before it is placed in a crucible. The recorded result can be present as below.

Weight of crucible + lid  = 23.70 g

Weight of crucible + lid + magnesium  =  24.90 g

Weight of crucible + lid after heating   =          25.70 g

[Relative atomic mass: Mg, 24; O, 16]

(a)       Define the term empirical and molecular formula. Give an appropriate example.

(b)        Calculate the mass of magnesium used in this experiment.

(c)      Find the mass of oxygen react with magnesium.

(d)       Why does magnesium ribbon is cleaned with the sand paper before the experiment.

(e)       Find the mole of magnesium used in this experiment.

(f)      Find the mole of oxygen that reacts with magnesium.

(g)      Suggest an empirical formulae for magnesium oxide.

(h)        State one observation from this experiment.

(i)         Why the lid has to be closed while doing the experiment?

(k)       The lid of the crucible has to be opened and closed quickly during the heating of

magnesium. Explain why?

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a. Empirical formula is chemical formula that show the simplest whole number ratio of atom of element in a compound. For a example, the empirical formula of glucose is CH2O.Molecular formula is chemical formula that show the actual number of atoms of each element present in one molecule of the compound. For a example, molecular formula of glucose is C6H12O6.
b. Mass of magnesium=24.9g-23.7=1.2g
c. Mass of oxygen=25.7g-24.9g=0.8g
d. To remove the oxide layer on the surface of magnesium ribbon.
e. Mole of magnesium=1.2÷24=0.05 mol
f. Mole of oxygen=0.8÷16=0.05mol
g. MgO
h. White fumes are produced when the magnesium ribbon start to burn.
I. To prevent the white fumes escape.
k. To let the oxygen into the crucible for the burning of magnesium

# data[].attributes.body
a)Emprical formula is a chemical formula that show the simplest whole number ratio of atom of element in a compound, the molecular formula is a chemical formula that show the actual number of atoms of each element present in one molecule of the compound. The emprical formula of glucose is CH₂O, the molecular formula of glucose is C6H12O6.
b) the mass of magnesium=24.90g-23.70g=1.2g
c) the mass of oxygen=25.70g-24.90g=0.8g
d)To remove the oxygen layer on the surface of magnesium ribbon.
e) mole of magnesium=1.2÷24=0.5mol
f) mole of oxygen=0.8÷16=0.5mol
g) MgO
h) White fumes are produced when the magnesium started to burn.
i) To prevent the white smoke of magnesium oxide from escaping into surroundings.
k)To allow the oxygen gas enter to the crucible for the complete combustion of magnesium.

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a) The empirical formula is a chemical formula shows the simpliest ratio of the simpliest ratio of the atoms of the elements that combine to form the compound. Example, empirical formula for ethanoic acid is CH20.The molecular formula is a chemical formula that shows the actual number of the atoms that combine to form the compound.For example, the molecular formula for ethanoic acid is C2H4O2.
b) 1.2 gram
c) 0.8 gram
d) to remove the oxide layer on the surface of the magnesium.
e) 0.05 mol
f) 0.05 mol
g) MgO
h) white fumes will be produced when magnesium started to burn.
i) to prevent white fumes from escaping into surrounding.
j) to allow oxygen to enter the crucible and to support the combustion of magnesium.

# data[].attributes.body
A) Empirical formula is chemical formula that show the simplest whole number  ratio of atom of element in a compound. For an example ,the empirical formula of ethane is CH3. The molecular formula of the compound shows the actual numbers of the atoms of the element that combine to form the compound. For an example, the molecular formula of ethane is C2H6
B) The mass of Mg = 1.2g
C) The mass of O = 0.8g
D) To remove the oxide layer on the surface of magnesium.
E) The mol of Mg = 0.05mol
F) The mol of O = 0.05mol
G) Empirical formula = MgO
H) White fumes are produced when the magnesium ribbon start to burn.
I) To prevent the white fumes from escaping into surrounding.
K) To allow oxygen gas enter to the crucible for the complete combustion.

# data[].attributes.body
(a)Empirical formula is the chemical formula that show the                simplest whole number ratio of atom of element in a  compound.For an example, the empirical formula of glucose is CH2O.Molecular formula is the chemical formula that shows the actual number of atoms of each element present in one molecule of the compound.For an example, the molecular formula of glucose is C6H12O6.

(b)Mass of magnesium=24.90g-23.70g
=1.20g

(c)Mass of oxygen=25.70g-24.90g
=0.80g

(d)To remove the oxide layer on the surface of magnesium.

(e)Mol of magnesium=0.05mol

(f)Mol of oxygen =0.05mol

(g)Empirical formulae of magnesium oxide is MgO.

(h)White fumes are produced when the magnesium ribbon started to burn.

(i)To prevent the white fumes of magnesium oxide from escaping into surroundings.

(k)To allow the oxygen gas enter to the crucible for the complete combustion of magnesium.

# data[].attributes.body
1.  a)The empirical formula is a chemical formula that show the simplest whole number ratio of atoms of each element in a compound.The empirical formula of glucose is CH2O.The molecule formula is chemical formula that shows the actual number of atoms of each element present in one molecule of the compound.The molecular formula of glucose is C6H1206.
b) mass =24.90-23.70
=1.20g
c) mass =25.70-24.90
=0.80g
d) To clean the oxide layer on the surface of magnesium ribbon.
e) mole =1.20g/24
=0.05mol
f) mole =0.80g/16
=0.05mol
g) MgO
h) White fumes produced when the magnesium started to burn.
I) To prevent white smoke from escaping.
k) To allow the oxygen to enter crucible from outside for combustion of magnesium.

# data[].attributes.body
a. Empirical formula is the chemical formula that shows the simplest ratio of atom of element in a compound . The empirical formula for glucose is CH20 . The molecular formula is the chemical formula that show the actual number of atom of each element present in one molecule of the compound .
b. Mass of Mg = 1.2 g
c. Mass of O = 0.8 g
d. To remove the oxide layer on the magnesium oxide .
e. Mol of Mg = 0.05mol
f. Mol of O = 0.05 mol
g. Empirical formula = MgO
h. white fumes are produced when the magnesium ribbon start to burn
I. to prevent the white fumes from escape to the surrounding
k. to allow the oxygen to enter the crucible for burning process

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limjinsing
a)  Empirical formula is the chemical formula that show the simplest ratio of atom of element in a compound.For example,the empirical formula of ethanoic acid is  CH2O.
Molecular formula is the chemical formula that show the actual number of atom  of each element present in one molecule of the compound.Example,the molecular formula of ethanoic acid is C2h4O2.

b)  Mass of magnesium=24.90g-23.70g =1.2g

c)  Mass of oxygen=25.7g-24.9g=0.8g

d)  To remove the layer of magnesium oxide in to get the desired chemical reaction.

e)  Mol of magnesium=1.2/24 =0.05mol

f)  Mol of oxygen=0.8/16 =0.05mol

g)  MgO.

h)  The white fumes will produce when the magnesium ribbon burned.

i)  To prevent the white fumes from escape to the surrounding.

k）To allow the oxygen to react with the magnesium and prevent the white fumes from escape out.

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LUI SHI YUN
a.)The empirical formula is the chemical formula that show the simplest whole number ratio of atom of element in a compound. The molecular formula is the chemical formula that show the actual number of atoms of each element present in one molecular of the compound.
b.)The mass of Mg=1.2g
c.)The mass of O=0.8g
d.)Remove the oxide layer on the surface
e.)Mol of Mg=1.2÷24=0.05mol
f.)Mol of O=0.8÷16=0.05mol
g.)MgO
h.)White fumes is produce when the magnesium ribbon start to burn
i.)To prevent the white fumes escape to surrounding
k.)Allow the oxygen enter the cricible during combustion

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A) The empirical formula shows the simplest whole number ratio of the atoms of the elements that combine to form the compound example :CH2O. The molecular formula shows the actual number of the atoms of elements that combine to form the compound example :C6H12O6.

B) 1.2g
C) 0.8g
D) To remove the oxide layer on the surface of magnesium ribbon.
E) 0.05mol
F) 0.05mol
G) MgO
H) The white fumes will produced when the magnesium is burning.
I) To ensure the white fumes can not escape from the crucible.
J) To allow the oxygen move in to the crucible for combustion magnesium ribbon.

# data[].attributes.body
a)Empirical formula is a chemical formula show the simplest whole number ratio of atom of element in a compound.For example,the empirical formula of glucose is CH2O and its simplify to CH2. Molecular formula is a chemical formula that shows the actual number of atoms of each element present in one molecule of the compound.For example,the molecular formula of glucose is C3H6 and its change to actual number of atoms to C3H6.
b)Mass of magnesium=24.90g-23.70g =1.2g
c)Mass of oxygen=25.7g-24.9g=0.8g
d)To remove the layer of magnesium oxide in to get the desired chemical reaction.
e)Mol of magnesium=1.2/24 =0.05mol
f)Mol of oxygen=0.8/16 =0.05mol
g)MgO
h)The white fumes will produce when the magnesium ribbon burned.
i)The lid provides a source of heat so that the magnesium metal atoms can overcome their activation energy.
k)To allow oxygen enter into crucible for combustion.

# data[].attributes.body
a) -The empirical formula is a chemical formula shows the simpliest ratio of the simpliest ratio of the atoms of the elements that combine to form the compound. For example, the empirical formula for glucose is CH20.
-The molecular formula is a chemical formula that shows the actual number of the atoms that combine to form the compound.For example,the molecular formula for glucose is C6H12O6.
b) The mass of Mg= 1.2 g
c) The mass of O= 0.8 g
d) to remove the oxide layer on the magnesium.
e) The mol of Mg= 0.05 mol
f) The mol of O= 0.05 mol
g) Empirical formula= MgO
h) white fumes is produce when magnesium start to burn.
i) to prevent white fumes from escaping to surround.
j) to allow oxygen to enter to support the burning process.

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a. Empirical formular is the chemica formular that
show the  simplest whole number ratio of atom of element in a compound. Molecular formular is the chemical formular that show the actual number of atoms of each element present in one molecule of the compound. Example of empirical formular of Glucose is CH2O , the example of molecular formular of Glucose is C6H12O6.
b. Mass of Mg= 24.9 - 23.7= 1.2g
c. Mass of O= 25.7 - 24.9= 0.8g
d. to remove the oxygen layer on the surface of magnesium.
e. Mole of Mg= 1.2÷24= 0.5mol
f. Mole of O= 0.8÷16= 0.5mol
g. MgO
h. Magnesium ribbon burn brightly when henting
I. To prevent the white smoke of magnesium oxide from escaping into surrounding.
j. To allow oxygen gas enter to the crucible for the complete combustion of magnesium.

# data[].attributes.body
Empirical formula is chemical formula that show the simplest whole number ratio of atom of element in a compound. Molecular formula is chemical formula that shows the actual number of atoms of each element present in one molecule of the compound. Example of empirical formula is glucose which has CH2O and example of molecular formula is glucose which has C6H12O6

1.2 gram
0.8 gram
This is to remove the oxide layer on the magnesium ribbon
0.05 mol
0.05 mol
MgO
White fumes will appear and produce when the magnesium ribbon starts to burn
This is to prevent the white fumes excape to the surrounding
To allow oxygen enter the crucible for complete combustion of magnesium

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The empirical formula is
the chemical formula that shows the simplest ratio
of atom for element in a compound. Molecular

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(a) The empirical formula is a chemical formula  shows the simplest ratio of the atoms of the elements that combine to form the compound. For example, C4H8. The molecular formula is a chemical formula shows the actual number of the atoms that combine to form the compound. For example, CH2.
(b) Mass of Mg = 24.90-23.70 = 1.20g
(c) Mass of O = 25.70-24.90 = 0.80g
(d) To remove the oxide layer on the surface of  magnesium.
(e) 1.2÷24 = 0.05mol
(f) 0.8÷16 = 0.05mol
(g) MgO
(h) The white fumes is produced during combustion.
(i) To prevent the white fumes of magnesium oxide from escaping into surrounding.
(k) To allow the oxygen enter the crucible for the  complete combustion of magnesium.

# data[].attributes.body
a)Empirical formula is the chemical formula that show the simplest ratio of atom of element in a compound.For example,the empirical formula of ethanoic acid is  CH2O.
Molecular formula is the chemical formula that show the actual number of atom  of each element present in one molecule of the compound.For example,the molecular formula of ethanoic acid is C2h4O2.

b)mass of Mg =1.2g

c)mass of O=0.8

d) To remove the oxide layer on the magnesium ribbon

e)mol of Mg=0.05mol

f)mol of O=0.05mol

g)Empirical formula=MgO

h)The white fumes is produce when the magnesium ribbon started to burn

i)To prevent the white fumes from escaping to surrounding

k)To allow the oxygen to enter the cruci

# data[].attributes.body
1

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a) -  The empirical formula  is the chemical formula that show the simplest ratio of atom of element in a compound . For example , the empirical formula of butene is CH2 .

-  The molecular formula is the chemical formula that shows the actual number of atoms of each element present in one molecule of the compound . For example , the molecular formula of butene is C4H8 .

b) Mass of Mg  = 24.90 - 23.70 = 1.2 g

c) Mass of O    =  25.70 - 24.90 = 0.8 g

d)  To remove the oxide layer of the surface of the magnesium ribbon .

e) Mol of Mg =  0.05 mol

f)  Mol of O    =  0.05 mol

g) MgO

h) White fumes are produced when the magnesium ribbon start to burn .

i) To prevent the white fumes from escape to the surrounding

k) To let the oxygen gas to enter the crucible for burning

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Jannah 4s
1
a)The empirical formula is formula that shows the simplest ratio of the number of atoms for each element in the compound.The molecular formula is formula that give us the actual number of atoms of each element that are present in a molecule of the compound.Example is the empirical formula of ethene is CH2 and the molecular formula of ethene is (CH2)2=C2H4.
b)mass of mg=24.90-23.70
=1.2gram
c)mass of O2=25.70-24.90
=0.8gram
d)to remove the oxide layer on the surface of magnesium
e)mole of mg=1.2÷24
=0.05moles
f)mole of O2=0.8÷16
=0.05moles
g)MgO
h)white fumes is produced when the magnesium ribbon start to burn.
i)to prevent the white fumes escape from the crucible
k)to allowed the oxygen gas enter to the crucible for burning

# data[].attributes.body
1.  empirical formula is the chemical formula that show the simplest ratio of atom for element in a compound. the molecular formula is the chemical formula that show the actual number of atom in a compound.the empirical formula of glucose is CH2O. the molecular formula of glucose is C6H12O6.

b. 1.2gram
c. 0.8gram
d. to remove the oxide layer on the magnesium oxide
e. 0.05 mol
f. 0.05 mol
g. MgO
h. white fumes produced when the magnesium started to burn
i. to prevent the white fumes from escaping to surrounding
k. to allow the oxygen enter the crucible to support the burning process.

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Exercise1: Empirical formulae of magnesium oxide

An experiment was conducted in a laboratory to determine the empirical formulae of magnesium oxide by heating magnesium strongly in a crucible. Magnesium ribbon was cleaned with a sand paper and it is coiled loosely before it is placed in a crucible. The recorded result can be present as below.

Weight of crucible + lid  = 23.70 g

Weight of crucible + lid + magnesium  =  24.90 g

Weight of crucible + lid after heating   =          25.70 g

[Relative atomic mass: Mg, 24; O, 16]

(a)       Define the term empirical and molecular formula. Give an appropriate example.

(b)        Calculate the mass of magnesium used in this experiment.

(c)      Find the mass of oxygen react with magnesium.

(d)       Why does magnesium ribbon is cleaned with the sand paper before the experiment.

(e)       Find the mole of magnesium used in this experiment.

(f)      Find the mole of oxygen that reacts with magnesium.

(g)      Suggest an empirical formulae for magnesium oxide.

(h)        State one observation from this experiment.

(i)         Why the lid has to be closed while doing the experiment?

(k)       The lid of the crucible has to be opened and closed quickly during the heating of

magnesium. Explain why?