• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Determination of the Formula of a Compound

Extracts from this document...

Introduction

Determination of the Formula of a Compound Results Mass of crucible and lid 23.0899g Mass of crucible, lid and Mg 23.1535g Mass of crucible, lid and product 23.1894g Mass of Mg used = 0.0636g (Mass of crucible, lid and Mg 23.1535 - Mass of crucible and lid 23.0899) ...read more.

Middle

Mg mass used 0.0636 Ratio Mean Mass of O used 0.0359 Atomic Mass 24 Atomic Mass 16 =0.00265 =0.00224375 Divide the smallest by the largest (0.00265/0.00224375) gives the ratio number of mole. Mg: 1.18105 O: 0.08466 Conclusion Oxygen is a very good oxidizing agent. ...read more.

Conclusion

These results have not shown a yield exact (simple) ratio. So these calculations show my results not to be entirely accurate. There may be several causes for this. The most obvious cause may have been due to experimenters error. The crucible lid may have been lifted to often allowing too much oxygen to react with the magnesium. The atoms are rearranged - diatomic oxygen molecules are split apart so that one oxygen atom combines with one magnesium atom. ...read more.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our GCSE Classifying Materials section.

## Found what you're looking for?

• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related GCSE Classifying Materials essays

1. ## Identification of an unknown compound.

To determine which of the two compounds the unknown chemical is, a few drops of the unknown will be added to blue litmus paper. The change in colour of the litmus paper will indicate the identity of the unknown compound.

2. ## Relationship between mass of MgO and its formula

128 u O } 48/128 = 0.375 u C / u O 8 atoms C @ 12 u each = 96 u C 16 atoms O @ 16 u each = 256 u O } 96/256 = 0.375 u C/ u O To clear this up we use John Dalton's theory, which supports the law of constant composition.

1. ## The role of mass customization and postponement in global logistics

* designing and developing products and services (P&S) to meet those needs. * manufacturing and delivering the P&S. * supporting the customer throughout the P&S life cycle, possibly including disposal. Like all businesses it must achieve all this profitably and speedily. 4.4.7 The benefits of MC are: * greater customer satisfaction and repeat business due to a better fit between customer needs and the P&S offering.

2. ## Finding the empirical formula of aluminium chloride.

Hence, according to Fajan's rule of bond type, aluminium chloride exhibits covalent bonding (to a certain extent). The aluminium chloride may be represented by the following dot and cross diagram: Yet, as can be seen above, aluminium only has six electrons in its outer shell and has not yet attained the stability of its nearest noble gas element argon.

• Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to