# zHW: Hill Notation

An analysis of a compound shows it to be 63.94 mass% Carbon, 7.15 mass% Hydrogen, and the rest (28.91%) is Oxygen. What is the Hill Notation for this compound?

## Contents

- Problem Data
- Convert to Molar Units
- Approximate Atomic Representation
- Error Analysis
- Hill Formula

## Problem Data

The problem data is given as mass fractions of the atomic species.

    wC = 0.6394;
    wH = 0.0715;
    wO = 1 - wC - wH;

## Convert to Molar Units

    nC = wC/molweight('C');
    nH = wH/molweight('H');
    nO = wO/molweight('O');

## Approximate Atomic Representation

This is the trickiest part of the problem. Here we construct approximate ratios of carbon, hydrogen, and oxygen in the compound and put this into an atomic represenation using a Matlab structure.

    d = min([nC nH nO]);

    r = struct([]);
    r(1).C = round(nC/d);
    r(1).H = round(nH/d);
    r(1).O = round(nO/d);

## Error Analysis

Let's see if our simple approximation solved the problem. If not, we would need to do some more work to develop a better approximation.

    vC = r.C*molweight('C');
    vH = r.H*molweight('H');
    vO = r.O*molweight('O');

    vT = vC+vH+vO;

    fprintf('Approximation Errors\n');
    fprintf('    Carbon: %5.2f %%\n',100*(vC/vT-wC)/wC);
    fprintf('  Hydrogen: %5.2f %%\n',100*(vH/vT-wH)/wH);
    fprintf('    Oxygen: %5.2f %%\n',100*(vO/vT-wO)/wO);

    Approximation Errors
        Carbon:  0.52 %
      Hydrogen:  0.58 %
        Oxygen: -1.29 %

## Hill Formula

The error analysis shows the approximate atomic representation is an acceptable solution to the problem. So the last step is to construct the Hill Formula.

    s = hillformula(r);
    fprintf('\nFormula in Hill Notation = %s\n',s{:});

    Formula in Hill Notation = C3H4O