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The relative formula mass of the compound MgCO_(x) is 84. Find x. A_(r) of Magnesium (Mg)=24 A_(r) of Carbon (C)=12 A_(r) of Oxygen (0)=16

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The relative formula mass of the compound
MgCO_(x) is 84. Find x.
A_(r) of Magnesium (Mg)=24
A_(r) of Carbon (C)=12
A_(r) of Oxygen (0)=16

The relative formula mass of the compound MgCO_(x) is 84. Find x. A_(r) of Magnesium (Mg)=24 A_(r) of Carbon (C)=12 A_(r) of Oxygen (0)=16

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DominicMaster · Tutor for 5 years

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<br />\[ x = 3 \]

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To solve for 'x', we have to understand the composition of the compound \(MgCO_x \). This compound consists of one Magnesium atom with an atomic mass of 24 amu, one Carbon atom with an atomic mass of 12 amu, and an unspecified 'x' number of Oxygen atoms, each with an atomic mass of 16 amu. The total relative formula mass was given to be 84. So, if we consider the formula masses of Magnesium and Carbon, we are left with an equation to find how many Oxygen atoms suffice the remaining mass to reach 84. Using those individual atomic masses, we can write:<br />\[<br />24(Mg) + 12 (C) + 16x (O) = 84<br />\]<br />Now we are going to isolate \(x\), multiplying 'x' by 16 inside parenthesis, applying distributive property.<br />\[<br />24 + 12 + 16x = 84<br />\]<br />Adding up 24 and 12 gives:<br />\[<br />36 + 16x = 84<br />\]<br />Subtract 36 from both sides:<br />\[<br />16x = 48<br />\]<br />Then we divide through by 16 to find the value of x:<br />\[<br />x = 48 / 16<br />\]<br />When we calculate, we got
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