text: # 300g of mg is burned on fire to produce magnesium oxide. final product weighs 4.97. what is the empirical formula for magnesium oxide?
  
  Mar 6, 2018
  
  On the (guessed) data we would plump for $M g O$...
  
  #### Explanation:
  
  The empirical formula is the simplest whole number ratio defining constituent atoms in a species... And so we interrogate the moles of magnesium and oxygen in the given problem.
  
  $\text{Moles of magnesium} = \frac{300 \times {10}^{-} 3 \cdot g}{24.3 \cdot g \cdot m o {l}^{-} 1} = 0.0123 \cdot m o l$
  
  $\text{Moles of oxygen} = \frac{\left(497 - 300\right) \times {10}^{-} 3 \cdot g}{16.0 \cdot g \cdot m o {l}^{-} 1} = 0.0123 \cdot m o l$
  
  And thus there are equimolar quantities of magnesium and oxygen in the given mass so we gets...an empirical formula of $M g O$..
  
  ...to do this formally we have...$M {g}_{\frac{0.0123 \cdot m o l}{0.0123 \cdot m o l}} {O}_{\frac{0.0123 \cdot m o l}{0.0123 \cdot m o l}} \equiv M g O$
  
  Mar 6, 2018
  
  MgO is Magenesium Oxide. 497.5 gms
  
  #### Explanation:
  
  $2 M g + {O}_{2} = 2 M g O$
  
  The molar mass of Magnesium is 24.31 gms/mol
  So 300 gms will be 12.34 mols
  
  Molar mass of MgO is 40.3 gms/mol
  
  So 12.34 mols would be 497.5 gms
explanation: The document contains a chemistry problem that involves calculating the empirical formula of magnesium oxide from given masses of magnesium and the final product. The text includes both the question and an answer, providing a detailed explanation of the calculation process.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # 300g of mg is burned on fire to produce magnesium oxide. final product weighs 4.97. what is the empirical formula for magnesium oxide?
  
  Mar 6, 2018
  
  On the (guessed) data we would plump for $M g O$...
  
  #### Explanation:
  
  The empirical formula is the simplest whole number ratio defining constituent atoms in a species... And so we interrogate the moles of magnesium and oxygen in the given problem.
  
  $\text{Moles of magnesium} = \frac{300 \times {10}^{-} 3 \cdot g}{24.3 \cdot g \cdot m o {l}^{-} 1} = 0.0123 \cdot m o l$
  
  $\text{Moles of oxygen} = \frac{\left(497 - 300\right) \times {10}^{-} 3 \cdot g}{16.0 \cdot g \cdot m o {l}^{-} 1} = 0.0123 \cdot m o l$
  
  And thus there are equimolar quantities of magnesium and oxygen in the given mass so we gets...an empirical formula of $M g O$..
  
  ...to do this formally we have...$M {g}_{\frac{0.0123 \cdot m o l}{0.0123 \cdot m o l}} {O}_{\frac{0.0123 \cdot m o l}{0.0123 \cdot m o l}} \equiv M g O$
  
  Mar 6, 2018
  
  MgO is Magenesium Oxide. 497.5 gms
  
  #### Explanation:
  
  $2 M g + {O}_{2} = 2 M g O$
  
  The molar mass of Magnesium is 24.31 gms/mol
  So 300 gms will be 12.34 mols
  
  Molar mass of MgO is 40.3 gms/mol
  
  So 12.34 mols would be 497.5 gms
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a chemistry problem that involves calculating the empirical formula of magnesium oxide from given masses of magnesium and the final product. The text includes both the question and an answer, providing a detailed explanation of the calculation process.
  
  Label: question-and-answer