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uranmaximum [27]
2 years ago
6

1. When conducting this experiment, some procedures call for heating the substance several times and recording the mass after ea

ch heating, continuing until the mass values are constant. Explain the purpose of this process and how it might reduce errors.
Chemistry
1 answer:
zvonat [6]2 years ago
3 0

The purpose of this process is to reduce the risk of having outliers of false results.

<h3> </h3><h3>What is a good scientific experiment?</h3>

It is important to bear in mind that a good scientific experiment usually tests one independent variable at a time, and that is why it is important to keep all other potential variables constant and to use a control group.

Although, the experiment you are referring to in your question is missing the general principle when conducting an experiment is the running of multiple tests before coming up with an actual test result.

Therefore, when completing a science experiment it is important to run multiple tests like heating the substance several times until a constant mass value is obtained,  in other to ascertain the results. the reason why such process is undertaken is to reduce the risk of having outliers of false results.  

A scientific experiment involves several procedures and this depends on the type and purpose of the experiment.

for example ;

An experiment with the purpose of determining how the mass of an object reacts to heat will require the mass of the object to be recorded  after heating the object/substance several times.

Hence we can conclude that the purpose of the process is to reduce the risk of having outliers of false results.

learn more about Scientific experiment at : brainly.com/question/13184032

#SPJ1

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Answer:

1. Percent composition of  Al = 13.423 %

2.

  • Percent composition of Zn = 28.02 %
  • Percent composition of Cl = 30.6 %
  • Percent composition of O = 41.3 %

3. The empirical formula is C₅O₁₆

4. Molecular Formula= P₄O₆

Explanation:

Part first :

Data Given

Formula of the Molecule = Al₂ (CrO₄)₃

% of Al₂ = ?

> First of all find the atomic masses of each component in a molecule

For Al₂ (CrO₄)₃ atomic masses are given below

Al = 27 g/mol

Cr = 52 g/mol

O = 16 g/mol

> Then find the total masses of each component

2 atoms of Al = 27 g/mol x 2

= 54 g/mol

3 atoms of Cr = 52 g/mol x 3

= 156 g/mol

12 atoms of O = 16 g/mol x 12

= 192 g/mol

> find total Molar Mass of Molecule:

Molar Mass of Al₂ (CrO₄)₃ = [27x2 + 52x3 + 16x12]

Molar Mass of Al₂ (CrO₄)₃ = 402

Now to find the mass percent of Al

Formula used to find the Mass percent of a component

Percent composition of  Al = mass of Al in Molecula / molar mass of Al₂(CrO₄)₃ x 100%

Put the values

Percent composition of  Al =  54 (g/mol) / 402 (g/mol) x 100%

Percent composition of  Al = 13.423 %

_______________________________________

Part 2

Data Given

Formula of the Molecule = Zn(ClO₃)₂

% Zn = ?

% Cl = ?

% O = ?

> First of all find the atomic masses of each component in a molecule

For Zn(ClO₃)₂ atomic masses are given below

Zn = 65 g/mol

Cl = 35.5 g/mol

O = 16 g/mol

> Then find the total masses of each component

1 atoms of Zn= 65 g/mol x 1

= 65 g/mol

2 atoms of Cl = 35.5 g/mol x  

= 71 g/mol

6 atoms of O = 16 g/mol x 6

= 96 g/mol

> find total Molar Mass of Molecule:

Molar Mass of Zn(ClO₃)₂ = [65x1 + 35.5x2 + 16x6]

Molar Mass of Zn(ClO₃)₂ = 232g/mol

Now to find the mass percent of of each component one by one

1.  Formula used to find the mass percent of Zn

Percent composition of  Zn= mass of Zn in Molecular / molar mass of Zn(ClO₃)₂ x 100%

Put the values

Percent composition of Zn = 65(g/mol) / 232 (g/mol) x 100%

Percent composition of Zn = 28.02 %

-------------------

2.  Formula used to find the mass percent of Cl

Percent composition of  Cl = mass of Cl in Molecular / molar mass of Zn(ClO₃)₂ x 100%

Put the values

Percent composition of Cl = 71 (g/mol) / 232 (g/mol) x 100%

Percent composition of Cl = 30.6 %

---------------------

3.  Formula used to find the mass percent of O

Percent composition of  O = mass of O in Molecular / molar mass of Zn(ClO₃)₂ x 100%

Put the values

Percent composition of O = 96 (g/mol) / 232 (g/mol) x 100%

Percent composition of O = 41.3 %

________________________________________

Part 3:

Data Given

Percentage of C = 27.3 %

Percentage of O = 72.7 %

Emperical Formula of the compound = ?

Solution:

So the compound has 27.3 % C and 72% O

First, find the mass of each of the elements in 100 g of the Compound.

C = 27.3 g

O = 72 g

Now find how many moles are there for each element in 100 g of compound

For this molar mass are required

That is

C = 12 g/mol

O = 16 g/mol

Formula Used

mole of C = mass of C / Molar mass of C

 mole of C = 27.3 / 12 g/mol

  mole of C = 2.275

Formula Used

mole of O = mass of O / Molar mass of O

 mole of O = 72g / 16 g/mol

  mole of O = 7.2

Divide each one by the smallest number of moles

C = 2.275 / 2.275

C = 1

O = 7.2 / 2.275

O = 3.2

Multiply the mole fraction to a number to get the whole number.

C = 1 x 5 = 5

O = 3.2 x 5 =  16

So, the empirical formula is C₅O₁₆

______________________________________

Part 4

Data Given

Percentage of P= 56.38 %

Percentage of O = 43.62%

Molar Mass = 219.9g

Molecular Formula of the compound = ?

Solution:

First, find the mass of each of the elements in 100 g of the Compound.

Mass of P= 56.38g

Mass of O = 43.62g

Now find how many moles are there for each element in 100 g of compound

find the moles in total compounds

Formula Used

mole of P = mass of  / Molar mass of P

 mole of P = 56.38 g / 31 g/mol

  mole of P = 1.818

Formula Used

mole of O = mass of O / Molar mass of O

 mole of O = 43. 62 / 16 g/mol

  mole of O = 2.7262

Now

first find the Emperical formula

Divide each one by the smallest number of moles

P = 1.818 /1.818

P= 1

for oxygen

O = 2.7262 / 1.818

O = 1.5

Multiply the mole fraction to a number to get the whole number.

P = 1 x 2 = 2

O = 1.5 x 2 =  3

So, the empirical formula is P₂O₃

Now  

Find molar mass of the empirical formula P₂O₃

2 (31) + 3 (16) = 62 + 48 = 110

Now find that how many empirical units are in a molecular unit.

(219.9 g/mol) / ( 110 g/mol) =  empirical units per molecular unit

empirical units per molecular unit = 1.999 =2

A here we get two empirical units in a molecular unit,

So the molecular formula is:

2 (P₂O₃) = P₄O₆

7 0
3 years ago
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