Answer
Verified
428.4k+ views
Hint: The empirical formula is created, the molecular formula for a compound can be resolute if the molar mass of the compound is identified. Basically calculate the mass of the empirical formula and divide the molar mass of the compound by the mass of the empirical formula to discover the ratio between the molecular formula and the empirical formula. Multiply all the atoms by this ratio to catch the molecular formula.
Complete step by step answer:
For appetizers, make assured that the chemical equation is balanced
\[2M{g_{(s)}} + {\text{ }}{O_{2(g)}} \to \;2Mg{O_{(s)}}\]
Sign that the response consumes \[2\] moles of magnesium and \[1\] mole of oxygen gas and produces \[2\] moles of magnesium oxide.
You can hence say that the number of moles of each chemical type that takes part in this reaction must gratify.
Currently, you can change grams to moles by using molar mass. In this case, you have
\[\begin{array}{*{20}{l}}
{{M_M}Mg = 24.3050{\text{ }}g{\text{ }}mo{l^{ - 1}}} \\
{{M_M}O2 = 32.0{\text{ }}g{\text{ }}mo{l^{ - 1}}} \\
{{M_M}MgO = 40.3050{\text{ }}g{\text{ }}mo{l^{ - 1}}}
\end{array}\]
So, change the mass of magnesium from grams to simply, use the element's molar mass as a renovation factor.
\[16.2g*{\text{ }}(1{\text{ }}mole{\text{ }}Mg{\text{ }}/{\text{ }}24.3050g)\]= \[0.6665\;\] moles Mg
Do the same for the example of oxygen gas.
\[10.67g*{\text{ }}(1{\text{ }}mole{\text{ }}{O_2}/{\text{ }}32.0g)\]= \[0.3334\;\] moles $O_2$
To conclude, do the similar for the magnesium oxide.
\[26.86g*{\text{ }}(1{\text{ }}mole{\text{ }}MgO{\text{ }}/40.3050g)\]= \[0.6664\;\] moles MgO
Thus , you see that when \[0.6665\;\] moles of magnesium react with \[0.3334\;\] moles of oxygen gas, you get \[0.6664\;\] moles of magnesium oxide.
The principles are not a strict match because of turning and of the values used for the molar masses of the three chemical types.
Note:
Stoichiometry agrees to make calculations about the outcomes of chemical reactions.
-Calculate the mass of a product of a chemical reaction if given the starting masses of reactants.
-Calculate the volume of a gas which will be produced by a reaction if given the first amounts of reactants.
-Define the optimum ratio of reactants for a chemical reaction so that all reactants are fully used.
Complete step by step answer:
For appetizers, make assured that the chemical equation is balanced
\[2M{g_{(s)}} + {\text{ }}{O_{2(g)}} \to \;2Mg{O_{(s)}}\]
Sign that the response consumes \[2\] moles of magnesium and \[1\] mole of oxygen gas and produces \[2\] moles of magnesium oxide.
You can hence say that the number of moles of each chemical type that takes part in this reaction must gratify.
moles Mg | moles $O_2$ | moles MgO |
2 | 1 | 2 |
Currently, you can change grams to moles by using molar mass. In this case, you have
\[\begin{array}{*{20}{l}}
{{M_M}Mg = 24.3050{\text{ }}g{\text{ }}mo{l^{ - 1}}} \\
{{M_M}O2 = 32.0{\text{ }}g{\text{ }}mo{l^{ - 1}}} \\
{{M_M}MgO = 40.3050{\text{ }}g{\text{ }}mo{l^{ - 1}}}
\end{array}\]
So, change the mass of magnesium from grams to simply, use the element's molar mass as a renovation factor.
\[16.2g*{\text{ }}(1{\text{ }}mole{\text{ }}Mg{\text{ }}/{\text{ }}24.3050g)\]= \[0.6665\;\] moles Mg
Do the same for the example of oxygen gas.
\[10.67g*{\text{ }}(1{\text{ }}mole{\text{ }}{O_2}/{\text{ }}32.0g)\]= \[0.3334\;\] moles $O_2$
To conclude, do the similar for the magnesium oxide.
\[26.86g*{\text{ }}(1{\text{ }}mole{\text{ }}MgO{\text{ }}/40.3050g)\]= \[0.6664\;\] moles MgO
Thus , you see that when \[0.6665\;\] moles of magnesium react with \[0.3334\;\] moles of oxygen gas, you get \[0.6664\;\] moles of magnesium oxide.
The principles are not a strict match because of turning and of the values used for the molar masses of the three chemical types.
Note:
Stoichiometry agrees to make calculations about the outcomes of chemical reactions.
-Calculate the mass of a product of a chemical reaction if given the starting masses of reactants.
-Calculate the volume of a gas which will be produced by a reaction if given the first amounts of reactants.
-Define the optimum ratio of reactants for a chemical reaction so that all reactants are fully used.
Recently Updated Pages
10 Examples of Evaporation in Daily Life with Explanations
10 Examples of Diffusion in Everyday Life
1 g of dry green algae absorb 47 times 10 3 moles of class 11 chemistry CBSE
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
What is the meaning of celestial class 10 social science CBSE
What causes groundwater depletion How can it be re class 10 chemistry CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
In the tincture of iodine which is solute and solv class 11 chemistry CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE