Taking into account the definition of a system of linear equations and molar mass, the molar masses of elements A and B are 26.778 [tex]\frac{g}{mole}[/tex] and 17.998[tex]\frac{g}{mole}[/tex] respectively.
System of linear equations
A system of linear equations is a set of two or more equations of the first degree, in which two or more unknowns are related.
Solving a system of equations consists of finding the value of each unknown so that all the equations of the system are satisfied. That is to say, the values of the unknowns must be sought, with which when replacing, they must give the solution proposed in both equations.
Definition of molar mass
The molar mass of substance is a property defined as its mass per unit quantity of substance, in other words, molar mass is the amount of mass that a substance contains in one mole.
The molar mass of a compound (also called Mass or Molecular Weight) is the sum of the molar mass of the elements that form it (whose value is found in the periodic table) multiplied by the number of times they appear in the compound.
Molar masses of elements A and B
Being the molar masses of the compounds:
- A₃B: [tex]\frac{17.7 g}{0.18 moles} =[/tex] 98.33 [tex]\frac{g}{mole}[/tex]
- AB₂: [tex]\frac{11.3 g}{0.18 moles} =[/tex] 62.77 [tex]\frac{g}{mole}[/tex]
and considering the definition of molar mass of a compound, the system of equations to be solved is
[tex]\left \{ {{3mA + mB=98.33\frac{g}{mole} } \atop {mA + 2mB=62.77\frac{g}{mole} }} \right.[/tex]
where mA and mB are the molar masses of elements A and B respectively.
There are several methods to solve a system of equations, it is decided to solve it using the substitution method, which consists of clearing one of the two variables in one of the equations of the system and substituting its value in the other equation.
Isolating the variable mB from the first equation:
mB= 98.33 [tex]\frac{g}{mole}[/tex] - 3mA
and substituting in the second equation, you get:
mA + 2× (98.33 [tex]\frac{g}{mole}[/tex] - 3mA)=62.77[tex]\frac{g}{mole}[/tex]
Solving:
mA + 2× 98.33 [tex]\frac{g}{mole}[/tex] - 2×3mA=62.77[tex]\frac{g}{mole}[/tex]
mA + 196.66 [tex]\frac{g}{mole}[/tex] - 6mA=62.77[tex]\frac{g}{mole}[/tex]
mA - 6mA=62.77[tex]\frac{g}{mole}[/tex] - 196.66 [tex]\frac{g}{mole}[/tex]
(-5)mA= - 133.89 [tex]\frac{g}{mole}[/tex]
mA= (- 133.89 [tex]\frac{g}{mole}[/tex] )÷ (-5)
mA= 26.778 [tex]\frac{g}{mole}[/tex]
Remember that mB= 98.33 [tex]\frac{g}{mole}[/tex] - 3mA, replacing the value of mA:
mB= 98.33 [tex]\frac{g}{mole}[/tex] - 3×(26.778 [tex]\frac{g}{mole}[/tex])
mB= 80.334 [tex]\frac{g}{mole}[/tex]
mB= 17.998 [tex]\frac{g}{mole}[/tex]
Finally, the molar masses of elements A and B are 26.778 [tex]\frac{g}{mole}[/tex] and 17.998[tex]\frac{g}{mole}[/tex] respectively.
Learn more about
system of equations:
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molar mass:
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